Trying to get $\dfrac{d}{dx}\left[\frac{20e^x}{(e^x+4)^2}\right]$.
After quotient rule: $$f'(x)=20\dfrac{e^x(e^{2x}+8e^x+16)-e^x(2e^x+8e^x)}{(e^x+4)^4}\\\\\\= 20\dfrac{e^{3x}+8e^{2x}+16e^x-2e^{2x}-8e^x}{(e^x+4)^4}\\\\=20\dfrac{e^{3x}+6e^{2x}+8e^x}{(e^x+4)^4}\\\\=\dfrac{20e^x(e^x+4)(e^x+2)}{(e^x+4)^4}\\\\=\dfrac{20e^x(e^x+2)}{(e^x+4)^3}$$
But Wolframalpha says it should be: $-\dfrac{20e^x(e^x-4)}{(e^x+4)^3}$
Where is my mistake?
$$ f'(x)=20\frac{e^x(e^{2x}+8e^x+16)-e^x(2e^{\color{red}{2x}}+8e^x)}{(e^x+4)^2} $$ You also do $e^x\cdot 8e^{x}=8e^x$, which is wrong.
On the other hand, if you consider $$ g(x)=\frac{x}{(x+4)^2} $$ you have $$ g'(x)=\frac{(x+4)^2-x\cdot 2(x+4)}{(x+4)^2}= \frac{x+4-2x}{(x+4)^3}=-\frac{x-4}{(x+4)^3} $$ Since $$ f(x)=20g(e^x) $$ you have $$ f'(x)=20g'(e^x)\cdot e^x=-\frac{20e^x(e^x-4)}{(e^x+4)^3} $$
Using: $$ \frac{d}{dx}(e^x+4)^2=2e^x(e^x+4) $$ we have $$ f'(x)=\frac{20e^x(e^x+4)^2-20\cdot 2e^x(e^x+4)}{(e^x+4)^4}=\frac{20 e^x(e^x+4)(e^x+4-2e^x)}{(e^x+4)^4}= $$ $$ =\frac{20e^x(-e^x+4)}{(e^x+4)^3} $$
