Stewart somehow still manages to write in a language I have never seen. Anyways I am trying to decipher his text and he claims that "Remembering that $y$ is a function of $x$ and using the chain rule, we have: $$\frac{d}{dx} (y^2) = \frac{d}{dy} (y^2) \frac{dy}{dx} = 2y \frac{dy}{dx}.$$ I am not sure why $y$ is a function of $x$ what that means or why I am using the chain rule but what I think he is stating with that line is that to find the derivative of $y^2$ with respect to $x$ we need to find the derivative of $y^2$ and multiply it by the derivative of $y$ with respect to $x$ and that equals $2y$. To my mind it should be $2y(2)$ which is $4y$.

  • $\begingroup$ He’s saying that you first find the derivative of $y^2$ with respect to y, which is $2y$, and then multiply it by $dy/dx$, which you don’t yet know, because you don’t know what function of $x$ $y$ is. You will eventually solve for $dy/dx$ in terms of $x$ and $y$. $\endgroup$ – Brian M. Scott Sep 24 '11 at 23:24
  • $\begingroup$ The unknown quantity $y$ is just some variable that depends on $\mathbf{x}$ is some unspecified way, so it automatically can be considered a function of $x$. For example, if $y=e^x$ then $dy^2/dx= 2e^{2x}$, which is most certainly not the same as $4e^x$. The reason we do implicit differentiation instead of saying $y=f(x)$ and then finding $df/dx$ is that sometimes we don't have an explicit $f$ to work with, e.g. we might have some equation like $F(y,x)=0$ on our hands and can't solve for $y$ explicitly in terms of $x$. $\endgroup$ – anon Sep 24 '11 at 23:26
  • $\begingroup$ I don't really understand what that means, finding the derivative of y^2 with respect to y. Is like just like finding the derivative of x^2 with respect to x? $\endgroup$ – toby yeats Sep 25 '11 at 0:23
  • $\begingroup$ Yes. ${}{}{}{}$ $\endgroup$ – joriki Sep 25 '11 at 0:35

"I am not sure why $y$ is a function of $x$ what that means or[...]"

The whole reason for the word "implicit" is that $y$ will be a function of $x$. That comes before you talk about differentiation. Consider an example: $$ \begin{align} x^2 + y^2 = 1 & & & (\text{implicit}) \\ \\ y = \pm\sqrt{1-x^2} & & &(\text{explicit}) \end{align} $$ The first equation above defines $y$ implicitly as a function of $x$.
The second equation above defines $y$ explicitly as a funciton of $x$.

You should understand that before thinking about implicit differentiation.

If $y = f(x)$ the $\dfrac{dy}{dx} = f'(x)$, and $\dfrac{d}{dx} y^{20} = \left(\dfrac{d}{dy} y^{20}\right)\cdot\left(\dfrac{dy}{dx}\right)$. That's an ordinary use of the chain rule, which says $$ \frac{dz}{dx} = \frac{dz}{dy}\cdot\frac{dy}{dx}, $$ and in this case $z=y^{20}$.

  • $\begingroup$ Just to be clear, the derivative would be (1)(20y) wouldn't it? My book is finding $x^2 + y^2$ and comes to that being equal to 2x + 2 y which then equals $-x/y$ which I don't quit understand. $\endgroup$ – toby yeats Sep 25 '11 at 15:16
  • $\begingroup$ @Jordan, $dy^{20}/dy = 20 y^{19}$. $\endgroup$ – rcollyer Sep 28 '11 at 4:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.