I have a question regarding the problem below...
Using implicit differentiation, what is the derivative of $x = e^{x y}$?
I believe this is using a chain rule. However, I am not sure if I need to use product rule also with xy? I am not sure because it is the exponent.
So, I found $g(x) = x y$ and $f(x) = e^x$ and apply this to $f'(g(x) \cdot g'(x))$ than I got $1 = e^x (x y \cdot y y')$.
Do I need to use the product rule with XY? Am I doing something wrong with this step so far?
$\endgroup$1 Answer
$\begingroup$You have the right idea. In this case, if you assume $y$ is a function of $x$, plus let $g(x) = xy$ and $f(x) = e^x$, your equation then becomes
$$x = e^{xy} = e^{g(x)} = f(g(x)) \tag{1}\label{eq1A}$$
If you implicitly differentiate \eqref{eq1A} wrt $x$, you get by using that $f'(x) = f(x)$ and the chain rule (plus the product rule when differentiating $g(x) = xy$) the following
$$\begin{equation}\begin{aligned} 1 & = f'(g(x))g'(x) \\ 1 & = f(g(x))g'(x) \\ 1 & = e^{xy}(y + xy') \end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Note your expression of $f'(g(x)\cdot g'(x))$ should actually be $f'(g(x))\cdot g'(x)$, like the second line in \eqref{eq2A}. As I stated earlier, you're correct you need to use the product rule for $xy$. However, as for your result of $1 = e^x (x y \cdot y y')$, I'm sure how you go that.
$\endgroup$
