Implicit differentiation essentially takes advantage of the chain rule \(\displaystyle f(g(x))'=g'(x)f'(g(x))\). I'm going to solve a different problem for you so that you can see the steps: \(\displaystyle e^{3y+10x-3}=ln(x)-12y^2\).
First, let's take the derivative:
\(\displaystyle (3y'+10)e^{3y+10x-3}=1/x-24y'y\)
Notice that the first term is in the form of \(\displaystyle e^{g(x)}\). The chain rule tells us that the derivative of such a function will be \(\displaystyle g'(x)e^{g(x)}\) since \(\displaystyle \frac{d}{dx}e^x = e^x\). In this case, \(\displaystyle g(x) = 3y+10x-3\). Since \(\displaystyle \frac{d}{dx}y = y'\), \(\displaystyle g'(x) = 3y'+10\).
From here, it's all about isolating \(\displaystyle y'\):
\(\displaystyle (3y'+10)e^{3y+10x-3}=1/x-24y'y \to\) \(\displaystyle 3y'e^{3y+10x-3}+24y'y=1/x-10e^{3y+10x-3} \to\) \(\displaystyle y'(3e^{3y+10x-3}+24y)=1/x-10e^{3y+10x-3} \to\) \(\displaystyle y'=\frac{1/x-10e^{3y+10x-3}}{3e^{3y+10x-3}+24y}\)
And it could perhaps be simplified more, but this is the solution for \(\displaystyle y'\) in terms of \(\displaystyle x\) and y(x). Provided there were an explicit solution for y (i.e., \(\displaystyle y=f(x)\)), one could write \(\displaystyle y'\) in terms of only \(\displaystyle x\).
Anyway. Good luck solving your problem. Feel free to post your solution here for verification.
