Answer
$y' = \frac{(2-x)^4(3x^2-55x-52)}{2(x+3)^{8}~\sqrt{x+1}}$
Work Step by Step
$y = \frac{\sqrt{x+1}(2-x)^5}{(x+3)^7}$
We can find $y'$:
$y' = \frac{[\frac{1}{2\sqrt{x+1}}~(2-x)^5+5(2-x)^4(-1)~\sqrt{x+1}]~(x+3)^7-7(x+3)^6~\sqrt{x+1}(2-x)^5}{(x+3)^{14}}$
$y' = \frac{[\frac{1}{2\sqrt{x+1}}~(2-x)^5+5(2-x)^4(-1)~\sqrt{x+1}]~(x+3)^7-7(x+3)^6~\sqrt{x+1}(2-x)^5}{(x+3)^{14}}\cdot \frac{\sqrt{x+1}}{\sqrt{x+1}}$
$y' = \frac{[\frac{1}{2}~(2-x)^5+5(2-x)^4(-1)~(x+1)]~(x+3)^7-7(x+3)^6~(x+1)(2-x)^5}{(x+3)^{14}~\sqrt{x+1}}$
$y' = \frac{[(2-x)-10~(x+1)]~(x+3)-14~(x+1)(2-x)}{2(x+3)^{8}~\sqrt{x+1}}\cdot (2-x)^4$
$y' = \frac{(-11x-8)~(x+3)-14~(x+1)(2-x)}{2(x+3)^{8}~\sqrt{x+1}}\cdot (2-x)^4$
$y' = \frac{(-11x^2-41x-24)+(14x^2-14x-28)}{2(x+3)^{8}~\sqrt{x+1}}\cdot (2-x)^4$
$y' = \frac{3x^2-55x-52}{2(x+3)^{8}~\sqrt{x+1}}\cdot (2-x)^4$
$y' = \frac{(2-x)^4(3x^2-55x-52)}{2(x+3)^{8}~\sqrt{x+1}}$
