Find the derivative of $2^x-3^y=1$ and then find the limit of $dy/dx$ as $x\to\infty$ Please tell the answer
If $2^x-3^y=1$
then what is the value of 
$$\lim_{x\to\infty} \frac {dy}{dx}?$$
I have tried finding the derivative implicitly, but I only get $0$ on both sides.
 A: Write $y=f(x)$ (for some $f(x)$ that we don't know), then
$$
1=2^x-3^{f(x)}=\exp(x\log2)-\exp(f(x)\times\log(3))
$$
so differentiating both sides gives
$$
0=(\log 2)2^x-(\log 3)f'(x)3^{f(x)}=(\log 2)2^x-(\log 3)f'(x)(2^x-1).
$$
Simplify and take limit:
$$
f'(x)=\frac{2^x\log(2)}{(2^x-1)\log(3)}\to\frac{\log(2)}{\log(3)}\text{ as }x\to\infty.
$$
You can check this answer by noting that for very large $x$, $2^x\approx 3^y\iff x\log 2\approx y\log 3$ so for large $x$, $y$ varies linearly with $x$ with slope $\frac{\log(2)}{\log(3)}$.
A: In order to implicitly derive $f(x) + g(y) = C$, you must first decide what to differentiate by. In your case, I assume you are differentiating along $x$, meaning that $y$ is actually a function of $x$ and the equation is acually $$f(x) + g(y(x)) = C$$
In that case, the derivation should result in:
$$f'(x) + (g(y(x)))' = 0$$
It is simple to find $f'(x)$, even in your case. To find $g(y(x))$, you will have to find the derivative using the chain rule.
A: differencing $2^x - 3^y = 1$, we get $$d\left(2^x - 3^y\right) = 0$$
$$ \ln(2) \, 2^x\, dx - \ln 3 \, 3^y \, dy = 0 $$
$$\begin{align}\frac{dy}{dx} &= \frac{2^x\ln 2}{3^y\ln 3}\\ &= \frac{2^x\ln 2}{(2^x - 1)\ln 2}\rightarrow \frac{\ln 2}{\ln 3} \text{ as } x \to \infty.\end{align}$$
