1.Find a pair of function $f(x)$ and$g(x)$ such that$\frac{d(f(x))}{dx}= ln (2)*2^x*cos (g(x))$.
I can't figure out how to begin to solve this question. I know that there should be multiple pairs of solutions just don't know how to rearrange this equation to solve for them.
I'm not sure if this is right?
$f(x)=2^x\cos(g(x))\ln(2)dx$
2.Given$f(t) =t+ ln(t)$, find $\frac{d(f^{-1})}{dt}$ at $t= 1 +e$. Exact values. Note that the symbol$\frac{d(f^{-1})}{dt}$means the derivative of the inverse function with respect to t.
I've solved for the inverse of $f$ but have gotten stuck at $e^yy=e^x $
so I don't know how I can apply this to find$\frac{d(f^{-1})}{dt}$ at $t= 1 +e$.
Best Answer
For the second one you can use $$f(f^{-1})(t) = t$$
$$\implies \frac{d}{dt}[f(f^{-1}(t))] = 1$$
$$\implies f'(f^{-1}(t)) \cdot \frac{d}{dt}[f^{-1}(t)] = 1$$
Use the fact that $f^{-1}(1+e) = e$, we have
$$\frac{d}{dt}[f^{-1}(t)] = \frac{1}{f'(e)}$$