Question

let (f(x)=-6x^{4}sqrt{x}+\frac{3}{x^{3}sqrt{x}}).
(f^{prime}(x)=)

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=-6x^{4}\sqrt{x}+\frac{3}{x^{3}\sqrt{x}}$ as $f(x)=-6x^{4}x^{\frac{1}{2}} + 3x^{-3}x^{-\frac{1}{2}}=-6x^{\frac{8 + 1}{2}}+3x^{\frac{-6 - 1}{2}}=-6x^{\frac{9}{2}}+3x^{-\frac{7}{2}}$.

Step2: Apply the power - rule for differentiation

The power - rule states that if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$.
For $y=-6x^{\frac{9}{2}}$, $y^\prime=-6\times\frac{9}{2}x^{\frac{9}{2}-1}=-27x^{\frac{7}{2}}$.
For $y = 3x^{-\frac{7}{2}}$, $y^\prime=3\times(-\frac{7}{2})x^{-\frac{7}{2}-1}=-\frac{21}{2}x^{-\frac{9}{2}}$.

Step3: Combine the derivatives

$f^\prime(x)=-27x^{\frac{7}{2}}-\frac{21}{2x^{\frac{9}{2}}}$.

Answer:

$-27x^{\frac{7}{2}}-\frac{21}{2x^{\frac{9}{2}}}$