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What is the purpose or proof behind chain rule?


Chain Rule applied to Trig Functionschain rule with manual substitutionchain rule or product ruleHelp understand chain rule derivativeThe chain rule problem with second compositeWhy is the chain rule applied to derivatives of trigonometric functions?Proof involving multivariable chain ruleChain rule to differentiate $sin ^2frac{x}{2}$Partial Derivative and Chain RuleDifferentiate without using chain rule in 5 steps













2












$begingroup$


For example, take a function $sin x$. The derivative of this function is $cos x$.



The chain rule states that $frac{d}{dx} (f(g(x)))$ is $frac{d}{dx} g(x) frac{d}{dx} (f(g(x)))$. Again going back to the example above, now instead of $sin x$ lets take $sin 2x$.



Differentiating it without chain rule, we get $cos 2x$. However, using chain rule, we get $2cos 2x$.



So now the problem is that I don't see the purpose behind the chain rule. Why should $sin 2x$ be $2cos 2x$?



Is there any proof behind this chain rule? I really need to know as I getting many questions wromg without using the chain rule.










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    " Why should $sin 2x;$ be $;2cos 2x$?" No, it isn't: its derivative is. Why? Because that's what we get from theorems or from the definition of derivative as limit. That's all. And yes: of course there is proof of the chain rule: any decent calculus book includes it.
    $endgroup$
    – DonAntonio
    50 mins ago












  • $begingroup$
    In your post, when you are 'differentiating without chain rule', you are differentiating $sin 2x$ with respect to $2x$, rather than with respect to $x$.
    $endgroup$
    – Minus One-Twelfth
    49 mins ago












  • $begingroup$
    @DonAntonio what i meant was derivative. I was just writing that in short. U shoukd be able to understand that as this whole post is about derivative
    $endgroup$
    – rash
    45 mins ago










  • $begingroup$
    @MinusOne-Twelfth whether i am taking with respect to 2x or x, the derivative value isnt the same and thats my confusion. For example derivative of $sin 2x$ where $piover 2$. Differentiating with respect to 2x is -1 & with respect to x is -2. Why?
    $endgroup$
    – rash
    41 mins ago








  • 1




    $begingroup$
    @littleO I didnt say it was correct. It is just my confusion of why should it not be like that and should be $2cos x$
    $endgroup$
    – rash
    40 mins ago


















2












$begingroup$


For example, take a function $sin x$. The derivative of this function is $cos x$.



The chain rule states that $frac{d}{dx} (f(g(x)))$ is $frac{d}{dx} g(x) frac{d}{dx} (f(g(x)))$. Again going back to the example above, now instead of $sin x$ lets take $sin 2x$.



Differentiating it without chain rule, we get $cos 2x$. However, using chain rule, we get $2cos 2x$.



So now the problem is that I don't see the purpose behind the chain rule. Why should $sin 2x$ be $2cos 2x$?



Is there any proof behind this chain rule? I really need to know as I getting many questions wromg without using the chain rule.










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    " Why should $sin 2x;$ be $;2cos 2x$?" No, it isn't: its derivative is. Why? Because that's what we get from theorems or from the definition of derivative as limit. That's all. And yes: of course there is proof of the chain rule: any decent calculus book includes it.
    $endgroup$
    – DonAntonio
    50 mins ago












  • $begingroup$
    In your post, when you are 'differentiating without chain rule', you are differentiating $sin 2x$ with respect to $2x$, rather than with respect to $x$.
    $endgroup$
    – Minus One-Twelfth
    49 mins ago












  • $begingroup$
    @DonAntonio what i meant was derivative. I was just writing that in short. U shoukd be able to understand that as this whole post is about derivative
    $endgroup$
    – rash
    45 mins ago










  • $begingroup$
    @MinusOne-Twelfth whether i am taking with respect to 2x or x, the derivative value isnt the same and thats my confusion. For example derivative of $sin 2x$ where $piover 2$. Differentiating with respect to 2x is -1 & with respect to x is -2. Why?
    $endgroup$
    – rash
    41 mins ago








  • 1




    $begingroup$
    @littleO I didnt say it was correct. It is just my confusion of why should it not be like that and should be $2cos x$
    $endgroup$
    – rash
    40 mins ago
















2












2








2


1



$begingroup$


For example, take a function $sin x$. The derivative of this function is $cos x$.



The chain rule states that $frac{d}{dx} (f(g(x)))$ is $frac{d}{dx} g(x) frac{d}{dx} (f(g(x)))$. Again going back to the example above, now instead of $sin x$ lets take $sin 2x$.



Differentiating it without chain rule, we get $cos 2x$. However, using chain rule, we get $2cos 2x$.



So now the problem is that I don't see the purpose behind the chain rule. Why should $sin 2x$ be $2cos 2x$?



Is there any proof behind this chain rule? I really need to know as I getting many questions wromg without using the chain rule.










share|cite|improve this question









$endgroup$




For example, take a function $sin x$. The derivative of this function is $cos x$.



The chain rule states that $frac{d}{dx} (f(g(x)))$ is $frac{d}{dx} g(x) frac{d}{dx} (f(g(x)))$. Again going back to the example above, now instead of $sin x$ lets take $sin 2x$.



Differentiating it without chain rule, we get $cos 2x$. However, using chain rule, we get $2cos 2x$.



So now the problem is that I don't see the purpose behind the chain rule. Why should $sin 2x$ be $2cos 2x$?



Is there any proof behind this chain rule? I really need to know as I getting many questions wromg without using the chain rule.







calculus derivatives soft-question






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 52 mins ago









rashrash

48814




48814








  • 1




    $begingroup$
    " Why should $sin 2x;$ be $;2cos 2x$?" No, it isn't: its derivative is. Why? Because that's what we get from theorems or from the definition of derivative as limit. That's all. And yes: of course there is proof of the chain rule: any decent calculus book includes it.
    $endgroup$
    – DonAntonio
    50 mins ago












  • $begingroup$
    In your post, when you are 'differentiating without chain rule', you are differentiating $sin 2x$ with respect to $2x$, rather than with respect to $x$.
    $endgroup$
    – Minus One-Twelfth
    49 mins ago












  • $begingroup$
    @DonAntonio what i meant was derivative. I was just writing that in short. U shoukd be able to understand that as this whole post is about derivative
    $endgroup$
    – rash
    45 mins ago










  • $begingroup$
    @MinusOne-Twelfth whether i am taking with respect to 2x or x, the derivative value isnt the same and thats my confusion. For example derivative of $sin 2x$ where $piover 2$. Differentiating with respect to 2x is -1 & with respect to x is -2. Why?
    $endgroup$
    – rash
    41 mins ago








  • 1




    $begingroup$
    @littleO I didnt say it was correct. It is just my confusion of why should it not be like that and should be $2cos x$
    $endgroup$
    – rash
    40 mins ago
















  • 1




    $begingroup$
    " Why should $sin 2x;$ be $;2cos 2x$?" No, it isn't: its derivative is. Why? Because that's what we get from theorems or from the definition of derivative as limit. That's all. And yes: of course there is proof of the chain rule: any decent calculus book includes it.
    $endgroup$
    – DonAntonio
    50 mins ago












  • $begingroup$
    In your post, when you are 'differentiating without chain rule', you are differentiating $sin 2x$ with respect to $2x$, rather than with respect to $x$.
    $endgroup$
    – Minus One-Twelfth
    49 mins ago












  • $begingroup$
    @DonAntonio what i meant was derivative. I was just writing that in short. U shoukd be able to understand that as this whole post is about derivative
    $endgroup$
    – rash
    45 mins ago










  • $begingroup$
    @MinusOne-Twelfth whether i am taking with respect to 2x or x, the derivative value isnt the same and thats my confusion. For example derivative of $sin 2x$ where $piover 2$. Differentiating with respect to 2x is -1 & with respect to x is -2. Why?
    $endgroup$
    – rash
    41 mins ago








  • 1




    $begingroup$
    @littleO I didnt say it was correct. It is just my confusion of why should it not be like that and should be $2cos x$
    $endgroup$
    – rash
    40 mins ago










1




1




$begingroup$
" Why should $sin 2x;$ be $;2cos 2x$?" No, it isn't: its derivative is. Why? Because that's what we get from theorems or from the definition of derivative as limit. That's all. And yes: of course there is proof of the chain rule: any decent calculus book includes it.
$endgroup$
– DonAntonio
50 mins ago






$begingroup$
" Why should $sin 2x;$ be $;2cos 2x$?" No, it isn't: its derivative is. Why? Because that's what we get from theorems or from the definition of derivative as limit. That's all. And yes: of course there is proof of the chain rule: any decent calculus book includes it.
$endgroup$
– DonAntonio
50 mins ago














$begingroup$
In your post, when you are 'differentiating without chain rule', you are differentiating $sin 2x$ with respect to $2x$, rather than with respect to $x$.
$endgroup$
– Minus One-Twelfth
49 mins ago






$begingroup$
In your post, when you are 'differentiating without chain rule', you are differentiating $sin 2x$ with respect to $2x$, rather than with respect to $x$.
$endgroup$
– Minus One-Twelfth
49 mins ago














$begingroup$
@DonAntonio what i meant was derivative. I was just writing that in short. U shoukd be able to understand that as this whole post is about derivative
$endgroup$
– rash
45 mins ago




$begingroup$
@DonAntonio what i meant was derivative. I was just writing that in short. U shoukd be able to understand that as this whole post is about derivative
$endgroup$
– rash
45 mins ago












$begingroup$
@MinusOne-Twelfth whether i am taking with respect to 2x or x, the derivative value isnt the same and thats my confusion. For example derivative of $sin 2x$ where $piover 2$. Differentiating with respect to 2x is -1 & with respect to x is -2. Why?
$endgroup$
– rash
41 mins ago






$begingroup$
@MinusOne-Twelfth whether i am taking with respect to 2x or x, the derivative value isnt the same and thats my confusion. For example derivative of $sin 2x$ where $piover 2$. Differentiating with respect to 2x is -1 & with respect to x is -2. Why?
$endgroup$
– rash
41 mins ago






1




1




$begingroup$
@littleO I didnt say it was correct. It is just my confusion of why should it not be like that and should be $2cos x$
$endgroup$
– rash
40 mins ago






$begingroup$
@littleO I didnt say it was correct. It is just my confusion of why should it not be like that and should be $2cos x$
$endgroup$
– rash
40 mins ago












3 Answers
3






active

oldest

votes


















1












$begingroup$

Visually, the derivative is the slope of the tangent line, and the derivative allows us to take a nonlinear function $f$ and approximate it locally with a linear function (that is, a function whose graph is a straight line). In other words, if we know the value of $f(x)$, we can approximate the value of $f$ at a nearby point $x + Delta x$ as follows:
$$
tag{1} f(x + Delta x) approx f(x) + f'(x) Delta x.
$$



Now suppose that $f(x) = g(h(x))$. Then we can approximate $f(x + Delta x)$ by using the above approximation twice, first with $h$ and then with $g$, as follows:
begin{align}
f(x + Delta x) &= g(h(x + Delta x)) \
&approx g(h(x) + h'(x) Delta x) \
&approx g(h(x)) + g'(h(x)) h'(x) Delta x.
end{align}

Comparing this result with equation (1), we see that
$$
f'(x) = g'(h(x)) h'(x).
$$



This is not yet a rigorous proof, but it shows how easy it is to discover the chain rule, and this derivation can be made into a rigorous proof without too much additional effort.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    When you said that the differentiation of $sin2x$ is $cos2x$, you didn't actually differentiate $sin 2x$ with respect to $x$, you differentiated it with respect to $2x$. Because the differentiation rule is that
    $$ frac{d}{dx}sin x = cos x$$



    so, only by replacing ALL $x$ in the formula above can you follow the same rule without breaking it, which is
    $$ frac{d}{d(2x)}sin 2x = cos 2x$$



    However, the question isn't asking you to find $frac{d}{d(2x)} sin 2x$, it is asking you to find $frac{d}{dx} sin 2x$. See the difference here?



    Since you differentiated the outer function,$f$, with respect to $2x$, you differentiated it with respect to the inner function because $g(x)=2x$.
    So you actually got $frac{df}{dg}=cos 2x$.
    To get from $frac{df}{dg}$ to $frac{df}{dx}$, you just need to multiply by $frac{dg}{dx}$ because:
    $$frac{df}{dg}timesfrac{dg}{dx} = frac{df}{dx} $$
    after cancelling out the $dg$.
    In this problem, $frac{dg}{dx} = frac{d}{dx}2x = 2$.
    That is why you have to mulitply a $2$ to your $cos 2x$.






    share|cite|improve this answer










    New contributor




    Carina Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    $endgroup$













    • $begingroup$
      Great explanation, but I have to mention that "derive" does not mean "differentiate". The words are not interchangable.
      $endgroup$
      – dbx
      18 mins ago










    • $begingroup$
      @dbx thanks for the catch! I've updated it.
      $endgroup$
      – Carina Chen
      16 mins ago



















    1












    $begingroup$

    This is a good question in my opinion. WHY is the chain rule right?
    My quick answer is that you are using the chain rule already without knowing it in the product rule, power rule, ect:
    $$
    frac{d}{dx}x^n = nx^{n-1}cdot frac{d}{dx}x = nx^{n-1}
    $$

    So when you differentiate $sin x$ you are actually doing $cos x cdot x' = cos x$.
    For a more detailed answer, lets look at the definition of the derivative.



    $$
    F'(x) = lim_{yrightarrow x}frac{F(x)-F(y)}{x-y}
    $$

    so let $F(x) = f(g(x))$ and what do we get?
    $$
    F'(x) = lim_{yrightarrow x}frac{f(g(x)) - f(g(y))}{x-y}
    $$

    which we can't evaluate. Let us assume that $g(x) ne g(y)$ when $x$ is 'close' to $y$, then we can multiply the whole thing by 1 to get the product of two derivatives:
    $$
    F'(x) = lim_{yrightarrow x}frac{f(g(x)) - f(g(y))}{g(x)-g(y)}cdot lim_{yrightarrow x} frac{g(x)-g(y)}{x-y} = f'(g(x))g'(x)
    $$

    where if we want to be picky we can consider $g(x)=g(y)$ too.



    (What follows is quite informal) The chain rule actually says something fundamental about composition. We can think of the function $g(x)$ as 'stretching' or 'shrinking' the domain of $f$. When we differentiate we are differentiating with respect to $f$ under an 'unstretched' domain and must correct for our error by multiplying by the derivative of $g$ which is a measure of how severely the domain was stretched. This is why the power rule ect. do not seem to use the chain rule, the domain is unstretched, so our derivative doesn't need to be corrected at all!



    For your example of $sin 2x$ lets think about what is going on, we are essentially squeezing $sin x$ in the $x$ direction. But this will make the slope of the sine function increase in a predictable way, in fact the slope at every point of this squeezed graph is twice as big as the original sine graph, exactly as predicted by the chain rule!



    For more complicated $g(x)$ the chain rule measures the rate at which the domain is changing from $x$ at every point to make the derivative correct.






    share|cite|improve this answer











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Visually, the derivative is the slope of the tangent line, and the derivative allows us to take a nonlinear function $f$ and approximate it locally with a linear function (that is, a function whose graph is a straight line). In other words, if we know the value of $f(x)$, we can approximate the value of $f$ at a nearby point $x + Delta x$ as follows:
      $$
      tag{1} f(x + Delta x) approx f(x) + f'(x) Delta x.
      $$



      Now suppose that $f(x) = g(h(x))$. Then we can approximate $f(x + Delta x)$ by using the above approximation twice, first with $h$ and then with $g$, as follows:
      begin{align}
      f(x + Delta x) &= g(h(x + Delta x)) \
      &approx g(h(x) + h'(x) Delta x) \
      &approx g(h(x)) + g'(h(x)) h'(x) Delta x.
      end{align}

      Comparing this result with equation (1), we see that
      $$
      f'(x) = g'(h(x)) h'(x).
      $$



      This is not yet a rigorous proof, but it shows how easy it is to discover the chain rule, and this derivation can be made into a rigorous proof without too much additional effort.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Visually, the derivative is the slope of the tangent line, and the derivative allows us to take a nonlinear function $f$ and approximate it locally with a linear function (that is, a function whose graph is a straight line). In other words, if we know the value of $f(x)$, we can approximate the value of $f$ at a nearby point $x + Delta x$ as follows:
        $$
        tag{1} f(x + Delta x) approx f(x) + f'(x) Delta x.
        $$



        Now suppose that $f(x) = g(h(x))$. Then we can approximate $f(x + Delta x)$ by using the above approximation twice, first with $h$ and then with $g$, as follows:
        begin{align}
        f(x + Delta x) &= g(h(x + Delta x)) \
        &approx g(h(x) + h'(x) Delta x) \
        &approx g(h(x)) + g'(h(x)) h'(x) Delta x.
        end{align}

        Comparing this result with equation (1), we see that
        $$
        f'(x) = g'(h(x)) h'(x).
        $$



        This is not yet a rigorous proof, but it shows how easy it is to discover the chain rule, and this derivation can be made into a rigorous proof without too much additional effort.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Visually, the derivative is the slope of the tangent line, and the derivative allows us to take a nonlinear function $f$ and approximate it locally with a linear function (that is, a function whose graph is a straight line). In other words, if we know the value of $f(x)$, we can approximate the value of $f$ at a nearby point $x + Delta x$ as follows:
          $$
          tag{1} f(x + Delta x) approx f(x) + f'(x) Delta x.
          $$



          Now suppose that $f(x) = g(h(x))$. Then we can approximate $f(x + Delta x)$ by using the above approximation twice, first with $h$ and then with $g$, as follows:
          begin{align}
          f(x + Delta x) &= g(h(x + Delta x)) \
          &approx g(h(x) + h'(x) Delta x) \
          &approx g(h(x)) + g'(h(x)) h'(x) Delta x.
          end{align}

          Comparing this result with equation (1), we see that
          $$
          f'(x) = g'(h(x)) h'(x).
          $$



          This is not yet a rigorous proof, but it shows how easy it is to discover the chain rule, and this derivation can be made into a rigorous proof without too much additional effort.






          share|cite|improve this answer









          $endgroup$



          Visually, the derivative is the slope of the tangent line, and the derivative allows us to take a nonlinear function $f$ and approximate it locally with a linear function (that is, a function whose graph is a straight line). In other words, if we know the value of $f(x)$, we can approximate the value of $f$ at a nearby point $x + Delta x$ as follows:
          $$
          tag{1} f(x + Delta x) approx f(x) + f'(x) Delta x.
          $$



          Now suppose that $f(x) = g(h(x))$. Then we can approximate $f(x + Delta x)$ by using the above approximation twice, first with $h$ and then with $g$, as follows:
          begin{align}
          f(x + Delta x) &= g(h(x + Delta x)) \
          &approx g(h(x) + h'(x) Delta x) \
          &approx g(h(x)) + g'(h(x)) h'(x) Delta x.
          end{align}

          Comparing this result with equation (1), we see that
          $$
          f'(x) = g'(h(x)) h'(x).
          $$



          This is not yet a rigorous proof, but it shows how easy it is to discover the chain rule, and this derivation can be made into a rigorous proof without too much additional effort.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 24 mins ago









          littleOlittleO

          30k647110




          30k647110























              1












              $begingroup$

              When you said that the differentiation of $sin2x$ is $cos2x$, you didn't actually differentiate $sin 2x$ with respect to $x$, you differentiated it with respect to $2x$. Because the differentiation rule is that
              $$ frac{d}{dx}sin x = cos x$$



              so, only by replacing ALL $x$ in the formula above can you follow the same rule without breaking it, which is
              $$ frac{d}{d(2x)}sin 2x = cos 2x$$



              However, the question isn't asking you to find $frac{d}{d(2x)} sin 2x$, it is asking you to find $frac{d}{dx} sin 2x$. See the difference here?



              Since you differentiated the outer function,$f$, with respect to $2x$, you differentiated it with respect to the inner function because $g(x)=2x$.
              So you actually got $frac{df}{dg}=cos 2x$.
              To get from $frac{df}{dg}$ to $frac{df}{dx}$, you just need to multiply by $frac{dg}{dx}$ because:
              $$frac{df}{dg}timesfrac{dg}{dx} = frac{df}{dx} $$
              after cancelling out the $dg$.
              In this problem, $frac{dg}{dx} = frac{d}{dx}2x = 2$.
              That is why you have to mulitply a $2$ to your $cos 2x$.






              share|cite|improve this answer










              New contributor




              Carina Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$













              • $begingroup$
                Great explanation, but I have to mention that "derive" does not mean "differentiate". The words are not interchangable.
                $endgroup$
                – dbx
                18 mins ago










              • $begingroup$
                @dbx thanks for the catch! I've updated it.
                $endgroup$
                – Carina Chen
                16 mins ago
















              1












              $begingroup$

              When you said that the differentiation of $sin2x$ is $cos2x$, you didn't actually differentiate $sin 2x$ with respect to $x$, you differentiated it with respect to $2x$. Because the differentiation rule is that
              $$ frac{d}{dx}sin x = cos x$$



              so, only by replacing ALL $x$ in the formula above can you follow the same rule without breaking it, which is
              $$ frac{d}{d(2x)}sin 2x = cos 2x$$



              However, the question isn't asking you to find $frac{d}{d(2x)} sin 2x$, it is asking you to find $frac{d}{dx} sin 2x$. See the difference here?



              Since you differentiated the outer function,$f$, with respect to $2x$, you differentiated it with respect to the inner function because $g(x)=2x$.
              So you actually got $frac{df}{dg}=cos 2x$.
              To get from $frac{df}{dg}$ to $frac{df}{dx}$, you just need to multiply by $frac{dg}{dx}$ because:
              $$frac{df}{dg}timesfrac{dg}{dx} = frac{df}{dx} $$
              after cancelling out the $dg$.
              In this problem, $frac{dg}{dx} = frac{d}{dx}2x = 2$.
              That is why you have to mulitply a $2$ to your $cos 2x$.






              share|cite|improve this answer










              New contributor




              Carina Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$













              • $begingroup$
                Great explanation, but I have to mention that "derive" does not mean "differentiate". The words are not interchangable.
                $endgroup$
                – dbx
                18 mins ago










              • $begingroup$
                @dbx thanks for the catch! I've updated it.
                $endgroup$
                – Carina Chen
                16 mins ago














              1












              1








              1





              $begingroup$

              When you said that the differentiation of $sin2x$ is $cos2x$, you didn't actually differentiate $sin 2x$ with respect to $x$, you differentiated it with respect to $2x$. Because the differentiation rule is that
              $$ frac{d}{dx}sin x = cos x$$



              so, only by replacing ALL $x$ in the formula above can you follow the same rule without breaking it, which is
              $$ frac{d}{d(2x)}sin 2x = cos 2x$$



              However, the question isn't asking you to find $frac{d}{d(2x)} sin 2x$, it is asking you to find $frac{d}{dx} sin 2x$. See the difference here?



              Since you differentiated the outer function,$f$, with respect to $2x$, you differentiated it with respect to the inner function because $g(x)=2x$.
              So you actually got $frac{df}{dg}=cos 2x$.
              To get from $frac{df}{dg}$ to $frac{df}{dx}$, you just need to multiply by $frac{dg}{dx}$ because:
              $$frac{df}{dg}timesfrac{dg}{dx} = frac{df}{dx} $$
              after cancelling out the $dg$.
              In this problem, $frac{dg}{dx} = frac{d}{dx}2x = 2$.
              That is why you have to mulitply a $2$ to your $cos 2x$.






              share|cite|improve this answer










              New contributor




              Carina Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$



              When you said that the differentiation of $sin2x$ is $cos2x$, you didn't actually differentiate $sin 2x$ with respect to $x$, you differentiated it with respect to $2x$. Because the differentiation rule is that
              $$ frac{d}{dx}sin x = cos x$$



              so, only by replacing ALL $x$ in the formula above can you follow the same rule without breaking it, which is
              $$ frac{d}{d(2x)}sin 2x = cos 2x$$



              However, the question isn't asking you to find $frac{d}{d(2x)} sin 2x$, it is asking you to find $frac{d}{dx} sin 2x$. See the difference here?



              Since you differentiated the outer function,$f$, with respect to $2x$, you differentiated it with respect to the inner function because $g(x)=2x$.
              So you actually got $frac{df}{dg}=cos 2x$.
              To get from $frac{df}{dg}$ to $frac{df}{dx}$, you just need to multiply by $frac{dg}{dx}$ because:
              $$frac{df}{dg}timesfrac{dg}{dx} = frac{df}{dx} $$
              after cancelling out the $dg$.
              In this problem, $frac{dg}{dx} = frac{d}{dx}2x = 2$.
              That is why you have to mulitply a $2$ to your $cos 2x$.







              share|cite|improve this answer










              New contributor




              Carina Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.









              share|cite|improve this answer



              share|cite|improve this answer








              edited 17 mins ago





















              New contributor




              Carina Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.









              answered 23 mins ago









              Carina ChenCarina Chen

              113




              113




              New contributor




              Carina Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.





              New contributor





              Carina Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              Carina Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.












              • $begingroup$
                Great explanation, but I have to mention that "derive" does not mean "differentiate". The words are not interchangable.
                $endgroup$
                – dbx
                18 mins ago










              • $begingroup$
                @dbx thanks for the catch! I've updated it.
                $endgroup$
                – Carina Chen
                16 mins ago


















              • $begingroup$
                Great explanation, but I have to mention that "derive" does not mean "differentiate". The words are not interchangable.
                $endgroup$
                – dbx
                18 mins ago










              • $begingroup$
                @dbx thanks for the catch! I've updated it.
                $endgroup$
                – Carina Chen
                16 mins ago
















              $begingroup$
              Great explanation, but I have to mention that "derive" does not mean "differentiate". The words are not interchangable.
              $endgroup$
              – dbx
              18 mins ago




              $begingroup$
              Great explanation, but I have to mention that "derive" does not mean "differentiate". The words are not interchangable.
              $endgroup$
              – dbx
              18 mins ago












              $begingroup$
              @dbx thanks for the catch! I've updated it.
              $endgroup$
              – Carina Chen
              16 mins ago




              $begingroup$
              @dbx thanks for the catch! I've updated it.
              $endgroup$
              – Carina Chen
              16 mins ago











              1












              $begingroup$

              This is a good question in my opinion. WHY is the chain rule right?
              My quick answer is that you are using the chain rule already without knowing it in the product rule, power rule, ect:
              $$
              frac{d}{dx}x^n = nx^{n-1}cdot frac{d}{dx}x = nx^{n-1}
              $$

              So when you differentiate $sin x$ you are actually doing $cos x cdot x' = cos x$.
              For a more detailed answer, lets look at the definition of the derivative.



              $$
              F'(x) = lim_{yrightarrow x}frac{F(x)-F(y)}{x-y}
              $$

              so let $F(x) = f(g(x))$ and what do we get?
              $$
              F'(x) = lim_{yrightarrow x}frac{f(g(x)) - f(g(y))}{x-y}
              $$

              which we can't evaluate. Let us assume that $g(x) ne g(y)$ when $x$ is 'close' to $y$, then we can multiply the whole thing by 1 to get the product of two derivatives:
              $$
              F'(x) = lim_{yrightarrow x}frac{f(g(x)) - f(g(y))}{g(x)-g(y)}cdot lim_{yrightarrow x} frac{g(x)-g(y)}{x-y} = f'(g(x))g'(x)
              $$

              where if we want to be picky we can consider $g(x)=g(y)$ too.



              (What follows is quite informal) The chain rule actually says something fundamental about composition. We can think of the function $g(x)$ as 'stretching' or 'shrinking' the domain of $f$. When we differentiate we are differentiating with respect to $f$ under an 'unstretched' domain and must correct for our error by multiplying by the derivative of $g$ which is a measure of how severely the domain was stretched. This is why the power rule ect. do not seem to use the chain rule, the domain is unstretched, so our derivative doesn't need to be corrected at all!



              For your example of $sin 2x$ lets think about what is going on, we are essentially squeezing $sin x$ in the $x$ direction. But this will make the slope of the sine function increase in a predictable way, in fact the slope at every point of this squeezed graph is twice as big as the original sine graph, exactly as predicted by the chain rule!



              For more complicated $g(x)$ the chain rule measures the rate at which the domain is changing from $x$ at every point to make the derivative correct.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                This is a good question in my opinion. WHY is the chain rule right?
                My quick answer is that you are using the chain rule already without knowing it in the product rule, power rule, ect:
                $$
                frac{d}{dx}x^n = nx^{n-1}cdot frac{d}{dx}x = nx^{n-1}
                $$

                So when you differentiate $sin x$ you are actually doing $cos x cdot x' = cos x$.
                For a more detailed answer, lets look at the definition of the derivative.



                $$
                F'(x) = lim_{yrightarrow x}frac{F(x)-F(y)}{x-y}
                $$

                so let $F(x) = f(g(x))$ and what do we get?
                $$
                F'(x) = lim_{yrightarrow x}frac{f(g(x)) - f(g(y))}{x-y}
                $$

                which we can't evaluate. Let us assume that $g(x) ne g(y)$ when $x$ is 'close' to $y$, then we can multiply the whole thing by 1 to get the product of two derivatives:
                $$
                F'(x) = lim_{yrightarrow x}frac{f(g(x)) - f(g(y))}{g(x)-g(y)}cdot lim_{yrightarrow x} frac{g(x)-g(y)}{x-y} = f'(g(x))g'(x)
                $$

                where if we want to be picky we can consider $g(x)=g(y)$ too.



                (What follows is quite informal) The chain rule actually says something fundamental about composition. We can think of the function $g(x)$ as 'stretching' or 'shrinking' the domain of $f$. When we differentiate we are differentiating with respect to $f$ under an 'unstretched' domain and must correct for our error by multiplying by the derivative of $g$ which is a measure of how severely the domain was stretched. This is why the power rule ect. do not seem to use the chain rule, the domain is unstretched, so our derivative doesn't need to be corrected at all!



                For your example of $sin 2x$ lets think about what is going on, we are essentially squeezing $sin x$ in the $x$ direction. But this will make the slope of the sine function increase in a predictable way, in fact the slope at every point of this squeezed graph is twice as big as the original sine graph, exactly as predicted by the chain rule!



                For more complicated $g(x)$ the chain rule measures the rate at which the domain is changing from $x$ at every point to make the derivative correct.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  This is a good question in my opinion. WHY is the chain rule right?
                  My quick answer is that you are using the chain rule already without knowing it in the product rule, power rule, ect:
                  $$
                  frac{d}{dx}x^n = nx^{n-1}cdot frac{d}{dx}x = nx^{n-1}
                  $$

                  So when you differentiate $sin x$ you are actually doing $cos x cdot x' = cos x$.
                  For a more detailed answer, lets look at the definition of the derivative.



                  $$
                  F'(x) = lim_{yrightarrow x}frac{F(x)-F(y)}{x-y}
                  $$

                  so let $F(x) = f(g(x))$ and what do we get?
                  $$
                  F'(x) = lim_{yrightarrow x}frac{f(g(x)) - f(g(y))}{x-y}
                  $$

                  which we can't evaluate. Let us assume that $g(x) ne g(y)$ when $x$ is 'close' to $y$, then we can multiply the whole thing by 1 to get the product of two derivatives:
                  $$
                  F'(x) = lim_{yrightarrow x}frac{f(g(x)) - f(g(y))}{g(x)-g(y)}cdot lim_{yrightarrow x} frac{g(x)-g(y)}{x-y} = f'(g(x))g'(x)
                  $$

                  where if we want to be picky we can consider $g(x)=g(y)$ too.



                  (What follows is quite informal) The chain rule actually says something fundamental about composition. We can think of the function $g(x)$ as 'stretching' or 'shrinking' the domain of $f$. When we differentiate we are differentiating with respect to $f$ under an 'unstretched' domain and must correct for our error by multiplying by the derivative of $g$ which is a measure of how severely the domain was stretched. This is why the power rule ect. do not seem to use the chain rule, the domain is unstretched, so our derivative doesn't need to be corrected at all!



                  For your example of $sin 2x$ lets think about what is going on, we are essentially squeezing $sin x$ in the $x$ direction. But this will make the slope of the sine function increase in a predictable way, in fact the slope at every point of this squeezed graph is twice as big as the original sine graph, exactly as predicted by the chain rule!



                  For more complicated $g(x)$ the chain rule measures the rate at which the domain is changing from $x$ at every point to make the derivative correct.






                  share|cite|improve this answer











                  $endgroup$



                  This is a good question in my opinion. WHY is the chain rule right?
                  My quick answer is that you are using the chain rule already without knowing it in the product rule, power rule, ect:
                  $$
                  frac{d}{dx}x^n = nx^{n-1}cdot frac{d}{dx}x = nx^{n-1}
                  $$

                  So when you differentiate $sin x$ you are actually doing $cos x cdot x' = cos x$.
                  For a more detailed answer, lets look at the definition of the derivative.



                  $$
                  F'(x) = lim_{yrightarrow x}frac{F(x)-F(y)}{x-y}
                  $$

                  so let $F(x) = f(g(x))$ and what do we get?
                  $$
                  F'(x) = lim_{yrightarrow x}frac{f(g(x)) - f(g(y))}{x-y}
                  $$

                  which we can't evaluate. Let us assume that $g(x) ne g(y)$ when $x$ is 'close' to $y$, then we can multiply the whole thing by 1 to get the product of two derivatives:
                  $$
                  F'(x) = lim_{yrightarrow x}frac{f(g(x)) - f(g(y))}{g(x)-g(y)}cdot lim_{yrightarrow x} frac{g(x)-g(y)}{x-y} = f'(g(x))g'(x)
                  $$

                  where if we want to be picky we can consider $g(x)=g(y)$ too.



                  (What follows is quite informal) The chain rule actually says something fundamental about composition. We can think of the function $g(x)$ as 'stretching' or 'shrinking' the domain of $f$. When we differentiate we are differentiating with respect to $f$ under an 'unstretched' domain and must correct for our error by multiplying by the derivative of $g$ which is a measure of how severely the domain was stretched. This is why the power rule ect. do not seem to use the chain rule, the domain is unstretched, so our derivative doesn't need to be corrected at all!



                  For your example of $sin 2x$ lets think about what is going on, we are essentially squeezing $sin x$ in the $x$ direction. But this will make the slope of the sine function increase in a predictable way, in fact the slope at every point of this squeezed graph is twice as big as the original sine graph, exactly as predicted by the chain rule!



                  For more complicated $g(x)$ the chain rule measures the rate at which the domain is changing from $x$ at every point to make the derivative correct.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 12 mins ago

























                  answered 18 mins ago









                  Kyle CKyle C

                  314




                  314






























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