A curve has implicit equation $x2^y=\ln y$. Find an expression for $\frac{dy}{dx}$ in terms of $x$ and $y$.
I got
$$x 2^y \ln 2 \frac{dy}{dx}+2^y=\frac{1}{y}$$
$$\frac{dy}{dx}=\frac{\frac{1}{y}-2^y}{x \ln 2 \cdot 2^y}$$
which isn't the answer. Can someone show me how to do it?
I good idea is to realise that locally (near a point where $\displaystyle \frac{dy}{dx}\neq0$) we have $y=y(x)$ so that locally we have
$$\begin{align} x2^{y(x)}&=\ln (y(x)). \end{align}$$ This might better allow you to see the various chain rules. We differentiate both sides with respect to $x$, using a product rule and chain rules: $$ \begin{align} x\frac{d}{dx}2^{y(x)}+(1)2^{y(x)}&=\frac{1}{y(x)}\cdot\frac{d}{dx}y(x) \\ \Rightarrow x\ln 2\cdot 2^{y(x)}\cdot\frac{d}{dx}y(x)+2^{y(x)}&=\frac{1}{y(x)}\cdot\frac{d}{dx}y(x). \end{align}$$ Now all we can say about $\displaystyle \frac{d}{dx}y(x)$ is that locally it is $\displaystyle \frac{dy}{dx}$. Also, now that we have used it to see the chain rules, we can write $y(x)=y$ again: $$(\ln2)x2^y\frac{dy}{dx}+2^y=\frac{1}{y}\frac{dy}{dx}.$$
I trust that you can finish things from here.
I recommend reading about the implicit function theorem. Especially how it lets you calculate the derivatives of the implicit function using the original function. In this case $F(x,y)=x.2^y-\ln y$
