<s>[INST] Derivative with quotient rule 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?
 [/INST] $$
f'(x)=20\frac{e^x(e^{2x}+8e^x+16)-e^x(2e^{\color{red}{2x}}+8e^x)}{(e^x+4)^2}
$$
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
$$
f'(x)=20g'(e^x)\cdot e^x=-\frac{20e^x(e^x-4)}{(e^x+4)^3}
$$
</s>