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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 $4y$.

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  • $\begingroup$ 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
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"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 $z=y^{20}$.

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  • $\begingroup$ 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

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