Question

what is the value of $\frac{d}{dx}(sqrt5{x^{2}})$ at $x = 32$? you might need: calculator

Explanation:

Step1: Rewrite the function

Rewrite $\sqrt[5]{x^{2}}$ as $x^{\frac{2}{5}}$.

Step2: Apply power - rule for differentiation

The power - rule states that if $y = x^{n}$, then $\frac{dy}{dx}=nx^{n - 1}$. For $y=x^{\frac{2}{5}}$, we have $\frac{d}{dx}(x^{\frac{2}{5}})=\frac{2}{5}x^{\frac{2}{5}-1}=\frac{2}{5}x^{-\frac{3}{5}}$.

Step3: Evaluate at $x = 32$

Substitute $x = 32$ into $\frac{2}{5}x^{-\frac{3}{5}}$. Since $32=2^{5}$, then $x^{-\frac{3}{5}}=(2^{5})^{-\frac{3}{5}}=2^{-3}=\frac{1}{8}$. So $\frac{2}{5}x^{-\frac{3}{5}}\big|_{x = 32}=\frac{2}{5}\times\frac{1}{8}=\frac{1}{20}$.

Answer:

$\frac{1}{20}$