What is the derivative of $f (x) = {(\frac{x + 5}{x^{2} + 2})}^{2}$ $f(x) = ((x+5)/(x^2+2))^2$ ?

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$f (x) = {(\frac{x + 5}{x^{2} + 2})}^{2}$ $f(x) = ((x+5)/(x^2+2))^2$ ?

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Answer 1 · 2017-10-19T09:10:04

$f'(x)=-(2(x+5)(x^2+10x-2))/(x^2+2)^3$

$f(x)$ can be written as $f(x)=(x+5)^2/(x^2+2)^2=u/v$

The quotient rule says that if $f(x)=g(x)/(h(x))$ , then $f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2$

In this case:
$g(x)=(x+5)^2$
$g'(x)=[d/dx(x+5)*d/dx(u^2)],u=(x+5)$
$g'(x)=1*2u$
$g'(x)=1*2(x+5)$
$g'(x)=2(x+5)$

$h(x)=(x^2+2)^2$
$h'(x)=[d/dx(x^2+2)*d/dx(u^2)],u=(x^2+2)$
$h'(x)=2x*2u$
$h'(x)=2x*2(x^2+2)$
$h'(x)=4x(x^2+2)$

$f'(x)=((x^2+2)^(2)2(x+5)-(x+5)^(2)4x(x^2+2))/((x^2+2)^2)^2$

$f'(x)=(2(x^2+2)^2(x+5)-4x(x+5)^(2)(x^2+2))/(x^2+2)^4$

$f'(x)=(2(x^2+2)(x+5)-4x(x+5)^2)/(x^2+2)^3$

$f'(x)=(2(x+5)((x^2+2)-2x(x+5)))/(x^2+2)^3$

$f'(x)=(2(x+5)(-x^2-10x+2))/(x^2+2)^3$

$f'(x)=-(2(x+5)(x^2+10x-2))/(x^2+2)^3$