Question

differentiate.

$f(x)=8\sqrt{x}+7x^{2}+14$

$f(x)=$

Explanation:

Step1: Rewrite square - root as power

Rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$, so $f(x)=8x^{\frac{1}{2}} + 7x^{2}+14$.

Step2: Apply power rule

The power rule for differentiation is $\frac{d}{dx}(ax^{n})=nax^{n - 1}$.
For the first term: $\frac{d}{dx}(8x^{\frac{1}{2}})=8\times\frac{1}{2}x^{\frac{1}{2}-1}=4x^{-\frac{1}{2}}$.
For the second term: $\frac{d}{dx}(7x^{2})=2\times7x^{2 - 1}=14x$.
For the third term: $\frac{d}{dx}(14)=0$ since the derivative of a constant is 0.

Step3: Combine the derivatives

$f'(x)=4x^{-\frac{1}{2}}+14x + 0$.

Answer:

$f'(x)=4x^{-\frac{1}{2}}+14x$