Question

for the function $f(x)=2sqrt{2x^{5}+4e^{x}}$ find $f(x)$. answer: $f(x)=$

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=2\sqrt{2x^{5}+4e^{x}}$ as $f(x)=2(2x^{5}+4e^{x})^{\frac{1}{2}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = f(g(x))$, then $y'=f'(g(x))\cdot g'(x)$. Let $u = 2x^{5}+4e^{x}$, so $y = 2u^{\frac{1}{2}}$. First, find the derivative of $y$ with respect to $u$: $\frac{dy}{du}=2\times\frac{1}{2}u^{-\frac{1}{2}}=u^{-\frac{1}{2}}$. Then find the derivative of $u$ with respect to $x$: $\frac{du}{dx}=10x^{4}+4e^{x}$.

Step3: Calculate $f'(x)$

By the chain - rule $f'(x)=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $u = 2x^{5}+4e^{x}$, $\frac{dy}{du}=u^{-\frac{1}{2}}$ and $\frac{du}{dx}=10x^{4}+4e^{x}$ into the formula: $f'(x)=\frac{10x^{4}+4e^{x}}{\sqrt{2x^{5}+4e^{x}}}$.

Answer:

$\frac{10x^{4}+4e^{x}}{\sqrt{2x^{5}+4e^{x}}}$