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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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    Yes. ${}{}{}{}$2011-09-25

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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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    @Jordan, $dy^{20}/dy = 20 y^{19}$.2011-09-28