Question

differentiate.
f(x)=\frac{x^{4}e^{x}}{x^{4}+e^{x}}
f(x)= square
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Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = x^{4}e^{x}$ and $v=x^{4}+e^{x}$.

Step2: Find $u'$ using product - rule

The product - rule states that if $u = ab$ where $a = x^{4}$ and $b = e^{x}$, then $u'=a'b+ab'$. Since $a' = 4x^{3}$ and $b'=e^{x}$, we have $u'=4x^{3}e^{x}+x^{4}e^{x}$. And $v' = 4x^{3}+e^{x}$.

Step3: Substitute $u$, $v$, $u'$, $v'$ into quotient - rule

\[

$$\begin{align*} f'(x)&=\frac{(4x^{3}e^{x}+x^{4}e^{x})(x^{4}+e^{x})-(x^{4}e^{x})(4x^{3}+e^{x})}{(x^{4}+e^{x})^{2}}\\ &=\frac{4x^{3}e^{x}(x^{4}+e^{x})+x^{4}e^{x}(x^{4}+e^{x})-4x^{7}e^{x}-x^{4}e^{2x}}{(x^{4}+e^{x})^{2}}\\ &=\frac{4x^{7}e^{x}+4x^{3}e^{2x}+x^{8}e^{x}+x^{4}e^{2x}-4x^{7}e^{x}-x^{4}e^{2x}}{(x^{4}+e^{x})^{2}}\\ &=\frac{x^{8}e^{x}+4x^{3}e^{2x}}{(x^{4}+e^{x})^{2}} \end{align*}$$

\]

Answer:

$\frac{x^{8}e^{x}+4x^{3}e^{2x}}{(x^{4}+e^{x})^{2}}$