Help with function u of x (chain rule)

Quote:

Originally Posted by Yogi_Bear_79

$Given f(x) =\frac{(x^2 +3x +1)^5}{(x+3)^5}$ , identify a function of u of x and an integer n not equal to 1 such that $f(x)=u^n$ . Then compute $f'(x)$ .

You can redo this to,
$\left( \frac{x^2+3x+1}{x+3} \right)^5$
Therefore,
$u=\frac{x^2+3x+1}{x+3}$
Thus,
$\frac{du}{dx}=\frac{(x^2+3x+1)'(x+3)-(x^2+3x+1)(x+3)'}{(x+3)^2}$
Thus,
$\frac{du}{dx}=\frac{(2x+3)(x+3)-(x^2+3x+1)(1)}{(x+3)^2}$
Thus,
$\frac{du}{dx}=\frac{2x^2+9x+9-x^2-3x-1}{(x+3)^2}$
Thus,
$\frac{du}{dx}=\frac{x^2+6x+8}{x^2+6x+9}=\frac{x^2+ 6x+9-1}{x^2+6x+9}=1-(x+3)^{-2}$
And,
$\frac{dy}{du}=5u^4$
Thus,
$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=5\cdot \left( 1-(x+3)^{-2} \right) \cdot \frac{(x^2+3x+1)^4}{(x+3)^4}$
Thus,
$\frac{5(x^2+3x+1)}{(x+3)^4}-5(x^2+3x+1)(x+3)^2$