Solving this question using ln? I know that ln is used to bring down exponents usually, so I was wondering where would I go from there ? I've looked at symbolab and don't really understand the implicit differentiation approach.
Here's the question $e^{(x^2)y}=2x+2y$
I was able to get to:
$$x^2y \ln(e) = \ln(2x)+ \ln(2y)$$
Where would i go from here ?
 A: I'd like to point out that you have made a common mistake. You seem to believe (correct me if I'm wrong) that
$$\ln(a+b)=\ln a+\ln b$$
This is wrong. In actual fact, the log law that you may be confusing this with is:
$$\ln(ab)=\ln a+\ln b$$
Note: this log law is not unique to the natural logarithm, it is a property of all logarithms.
About your question, your equation has infinitely many solutions. I assume that you want to differentiate it implicitly (as opposed to solving it), so this is how you'd do it. It is inconvenient in this case to use $\ln$ as we don't have a product on the right hand side of our original equation, so I will differentiate directly:
$$e^{x^2y}=2x+2y$$
Now, differentiating with respect to $x$, and using the fact that $\frac{d}{dx}e^{f(x)}=f'(x)e^{f(x)}$ we have
$$\begin{align}
&\frac{d}{dx}(e^{x^2y})=\frac{d}{dx}(2x)+\frac{d}{dx}(2y)\\
&\left(2xy+x^2\frac{dy}{dx}\right)e^{x^2y}=2+2\frac{dy}{dx}
\end{align}$$
Now rearrange to find $\frac{dy}{dx}$. If you need any more help please don't hesitate to ask.
