I have the equation:
(x^2)(y^-3) + 3 = y
How do I go about implicitly differentiating this equation?
EDIT: Solve for dy/dx. Sorry about that.
Hi, Jeavus. You can probably achieve what this question wants of you, by differentiating both sides of the equation by $\displaystyle x$, assuming that $\displaystyle y$ is some function of $\displaystyle x$:
$\displaystyle x^2 y^{-3} + 3 = y$
$\displaystyle \Rightarrow \frac{d}{dx}\left[x^2 y^{-3} + 3\right] = \frac{d}{dx}\left[y\right] = \frac{dy}{dx}$
$\displaystyle \Rightarrow \left( 2x y^{-3} - 3 x^2 y^{-4} \frac{dy}{dx} \right) = \frac{dy}{dx}$
To do the last step, I used the "product rule" of differentiation.
If you rearrange this, you should be able to work out $\displaystyle \frac{dy}{dx}$ as a function of $\displaystyle x$ and $\displaystyle y$. Hope that helps!