$y^5 =(x+2)^4+(e^x)(ln y)−15$ finding $\frac{dy}{dx}$ at $(0,1)$ 
Unsure what to do regarding the $y^5$. 
Should I convert it to a $y$= function and take the $5$ root of the other side. Then differentiate? Any help would be great thanks.

Unfortunately we cannot rearrange this to find $y$. Fortunately we can still use implicit differentiation to find a derivative. Take the derivative of all terms w.r.t. $x$, using the chain rule and treating $y$ as a function of $x$ - we obtain
$$
5y^4\frac{\mathrm{d}y}{\mathrm{d}x} = 4(x+2)^3 + \mathrm{e}^x\ln y +\frac{\mathrm{e}^x}{y}\frac{\mathrm{d}y}{\mathrm{d}x}.
$$
Note that we used the product rule for the $\mathrm{e}^x\ln y$ term. Now the trick is to rearrange to solve for the derivative, and then substitute $(x,y) = (0,1)$. Can you take it from here? 
