How can I use the chain rule $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$ to differentiate an expression like $\frac{d}{dx}(y^3)$ ? The solution in the book is

\begin{equation*} \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}. \end{equation*}

I'm not sure how they got it though.

  • $\begingroup$ Wait... partial or total? One is different from the other... $\endgroup$ – J. M. is a poor mathematician Aug 20 '10 at 17:32
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    $\begingroup$ If it was $\partial(y^3)/\partial x$ isn't the result trivially 0? $\endgroup$ – kennytm Aug 20 '10 at 17:37
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    $\begingroup$ I believe total.. This is where I got the expression from en.wikibooks.org/wiki/Calculus/… $\endgroup$ – Tony Johnson Aug 20 '10 at 17:38
  • $\begingroup$ @Tony: Isn't the answer already in the link you've quoted...? $\endgroup$ – kennytm Aug 20 '10 at 17:39
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    $\begingroup$ Kenny: Exactly. Tony: In that case, you should always think of the other variables as functions of x; thus, you are in fact taking the derivative of $y(x)^3$. Then you apply the chain rule. $\endgroup$ – J. M. is a poor mathematician Aug 20 '10 at 17:51

In the problem, you want to differentiate the expression $\color{green}y^{\color{blue}3}$ with the variable $\color{red}x$. But there is no information on what $\color{green}y$ is. So we must assume $\color{green}y$ is an arbitrary function, i.e. $\color{green}y := \color{green}y(\color{red}x)$.

Now the crucial point is to note that the 3 itself is also a function, i.e. we can define $\color{blue}z(\color{green}y) = \color{green}y^{\color{blue}3}$.

In this way, we could apply the chain rule (some names changed to be consistent with the problem), $$ \frac{d \color{blue} z}{d \color{red} x} = \frac{d \color{blue} z}{d \color{green} y} \cdot \frac{d \color{green} y}{d \color{red} x} $$ i.e. explictly, $$ \frac{d \Bigl( \color{blue} z\bigl(\color{green}y(\color{red}x)\bigr) \Bigr)}{d \color{red} x} = \frac{d \Bigl( \color{blue} z \bigl(\color{green}y\color{silver}{(x)}\bigr) \Bigr) }{d\bigl( \color{green}y\color{silver}{(x)}\bigr)} \cdot \frac{d \bigl( \color{green} y (\color{red}x)\bigr) }{d \color{red} x} $$ In $d\color{blue}z/d\color{green}y$, since no other variables (like $\color{red}x$) is directly involved, we could treat $\color{green}y$ as a variable instead of an implicit function. In $d\color{green}y/d\color{red}x$, as the function $\color{green}y(\color{red}x)$ is arbitrary, the differentiated form is the best representation already. Therefore, the expression above simplifies to $$ \frac{d \left( \color{green}y^{\color{blue}3} \right) }{d \color{red} x} = \underbrace{ \color{teal}{ 3 \color{green}y^2 } }_{d(\color{green}y^{\color{blue}3})/d\color{green}y} \cdot \frac{d \color{green} y}{d \color{red} x} $$

  • $\begingroup$ Yea, in intro calculus books they typically leave out the exact definition of the functions... So I'm not sure I understand where you got them. Or how they apply to the operators in question. $\endgroup$ – Tony Johnson Aug 20 '10 at 17:40
  • $\begingroup$ Ok, I see. That helps a lot. Thanks! $\endgroup$ – Tony Johnson Aug 20 '10 at 18:42

I think there is a bit of a mix up over notation - specifically, the problem is using $y$ in two different contexts.

The definition of the chain rule that you need (and what you have correctly stated):

$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$

Now, for a composite function $y(u(x)) = [u(x)]^3$ (here, $y$ is the 'cubing' function), in order to take the derivative of $y$ with respect to $x$ we first take the derivative of $y$ w.r.t. $u$, then since $u$ is a function of $x$ (and should be written as $u(x)$ at this point) we use the chain rule as follows:

$\frac{dy}{du} = 3u^{2}$, and $\frac{du}{dx} = u'(x)$ since we don't have the expression of the function $u(x)$ itself. So it looks like $$\frac{d}{dx}([u(x)]^3) = 3[u(x)]^{2}u'(x).$$

This looks different because I have chosen to relabel the function $y^3$ as $u^{3}$ in order for the stated chain rule to 'look right.' Alternatively, we could state the chain rule as $\frac{dh}{dx} = \frac{dh}{dy}\frac{dy}{dx}$ where $h = h(y(x)) = [y(x)]^3$. Then $\frac{dh}{dy} = 3y^2$ and $\frac{dy}{dx}$ is just $y'$ again because we don't have a specific expression for that function. Here we would have $\frac{dh}{dx} = 3y^2y'$.

  • $\begingroup$ So if I understand you correctly, then when the book states that the answer is $\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}$, $y$ is a function of x (if understood implicitly) such that $y=u(x)$? $\endgroup$ – Tony Johnson Aug 20 '10 at 18:31
  • $\begingroup$ No, the book is correct with the $\frac{dy}{dx}$ at the end. The book is making its mistake (IMO) when it declares the chain rule for the derivative of a composite function $y$ w.r.t. $x$ and then asks for the derivative of $y^3$ w.r.t. $x$. In this terminology the function $y^3$ of the problem is the function $y$ in the statement of the chain rule. $\endgroup$ – Tom Stephens Aug 20 '10 at 18:37
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    $\begingroup$ After today you will be able to distinguish these things and make the proper substitutions without even thinking about it. The book is essentially teaching you an example of multiplication using $2 \cdot 2=4$ (so there can be confusion since $2+2=4$ as well...) $\endgroup$ – Tom Stephens Aug 20 '10 at 18:40
  • $\begingroup$ This helps also. Thanks! $\endgroup$ – Tony Johnson Aug 20 '10 at 18:43

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