Solving a Chain Rule Problem: Find Derivative of y

[efekwulsemmay]

Homework Statement

I need to find the derivative of:

y=\left(4x+3\right)^{4}\cdot\left(x+1\right)^{-3}

Homework Equations

Chain Rule
Quotient or Product Rule

The Attempt at a Solution

So I tried to use quotient rule because

\left(4x+3\right)^{4}\cdot\left(x+1\right)^{-3}=\frac{\left(4x+3\right)^{4}}{\left(x+1\right)^{3}}

thus by quotient rule

y=\frac{\left(4x+3\right)^{4}}{\left(x+1\right)^{3}}, \frac{dy}{dx}=\frac{\left[\left(x+1\right)^{3}\cdot4\left(4x+3\right)^{3}\cdot4\right]-\left[\left(4x+3\right)^{4}\cdot3\left(x+1\right)^{2}\cdot1\right]}{\left[\left(x+1\right)^{3}\right]^{2}}

=\frac{\left[16\left(4x+3\right)^{3}\cdot\left(x+1\right)^{3}\right]-\left[3\left(4x+3\right)^{4}\cdot\left(x+1\right)^{2}\right]}{\left(x+1\right)^{6}}

=\frac{16\left(4x+3\right)^{3}\cdot\left(x+1\right)^{3}}{\left(x+1\right)^{6}}-\frac{3\left(4x+3\right)^{4}\cdot\left(x+1\right)^{2}}{\left(x+1\right)^{6}}

=\frac{16\left(4x+3\right)^{3}}{\left(x+1\right)^{2}}-\frac{3\left(4x+3\right)^{4}}{\left(x+1\right)^{3}}

I don't know where to go from here... I know that the answer to the problem is

\frac{\left(4x+3\right)^{3}\left(4x+7\right)}{\left(x+1\right)^{4}}

I just don't know how the hell I am supposed to get there.

 

[LCKurtz]

Instead of breaking the fraction into two on the third line, factor everything you can in the numerator on the second line and see what happens.

 

[efekwulsemmay]

Good god I feel foolish. It was really that simple. The thought, "why didn't I think of that?" comes to mind. :) Thank you for your help.