# Homework Help: Chain Rule Problem

1. Sep 21, 2009

### efekwulsemmay

1. The problem statement, all variables and given/known data
I need to find the derivative of:

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

2. Relevant equations
Chain Rule
Quotient or Product Rule

3. 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.

2. Sep 21, 2009

### 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.

3. Sep 21, 2009

### 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.