Originally Posted by

**Scott H** The trick for the first problem is to look at both sides of the equation as functions of $\displaystyle x$ and differentiate them, using the Chain Rule for $\displaystyle y$.

This gives us

$\displaystyle \frac{d}{dx}\sin(2x + 3y) = \frac{d}{dx}(3x^3y^2 + 4)$

$\displaystyle \cos(2x + 3y)\cdot 3\frac{dy}{dx} = 9x^2y^2 + 3x^3\cdot 2y\frac{dy}{dx}$

$\displaystyle \frac{dy}{dx}(\cos(2x + 3y)\cdot3 - 3x^3\cdot2y) = 9x^2y^2$

$\displaystyle \frac{dy}{dx} = 3x^2y^2\cdot(\cos(2x + 3y) - 2x^3y)^{-1}.$

For the second problem, it is useful to ask what constants $\displaystyle A$ and $\displaystyle B$ could make the equation

$\displaystyle \frac{1}{(x + 5)(x - 5)} = \frac{A}{x + 5} + \frac{B}{x - 5}$

true. This way, the integrands are simplified. In finding $\displaystyle A$ and $\displaystyle B$, we use the fact that the equation must hold for *all* $\displaystyle x$, not just one value.