Question

find the derivative of the function.
y = 8\sqrt{x}+8x^{\frac{3}{4}}
\frac{dy}{dx}=\square

Explanation:

Step1: Rewrite the function

Rewrite $y = 8\sqrt{x}+8x^{\frac{3}{4}}$ as $y = 8x^{\frac{1}{2}}+8x^{\frac{3}{4}}$.

Step2: Apply the power - rule

The power - rule for differentiation is $\frac{d}{dx}(ax^n)=nax^{n - 1}$.
For the first term $8x^{\frac{1}{2}}$, using the power - rule: $\frac{d}{dx}(8x^{\frac{1}{2}})=\frac{1}{2}\times8x^{\frac{1}{2}-1}=4x^{-\frac{1}{2}}$.
For the second term $8x^{\frac{3}{4}}$, using the power - rule: $\frac{d}{dx}(8x^{\frac{3}{4}})=\frac{3}{4}\times8x^{\frac{3}{4}-1}=6x^{-\frac{1}{4}}$.

Step3: Find the derivative of the function

$y^\prime=\frac{dy}{dx}=\frac{d}{dx}(8x^{\frac{1}{2}})+\frac{d}{dx}(8x^{\frac{3}{4}})=4x^{-\frac{1}{2}} + 6x^{-\frac{1}{4}}$.

Answer:

$4x^{-\frac{1}{2}}+6x^{-\frac{1}{4}}$