Question

let $f(x)=6x^{4}sqrt{x}+\frac{-2}{x^{2}sqrt{x}}$.
hint: this derivative is easier to find if you first change the terms with radicals to fractional exponents, then simplify using basic exponent rules.
$f(x)=$
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Explanation:

Step1: Rewrite the function using fractional exponents

$$\begin{align*} f(x)&=6x^{4}\cdot x^{\frac{1}{2}}+\frac{-2}{x^{2}\cdot x^{\frac{1}{2}}}\\ &=6x^{4 + \frac{1}{2}}-2x^{-2-\frac{1}{2}}\\ &=6x^{\frac{9}{2}}-2x^{-\frac{5}{2}} \end{align*}$$

Step2: Apply the power rule for differentiation

The power rule is \((x^{n})^\prime=nx^{n - 1}\).
For the first term \(y = 6x^{\frac{9}{2}}\), its derivative \(y^\prime=6\times\frac{9}{2}x^{\frac{9}{2}-1}=27x^{\frac{7}{2}}\).
For the second term \(y=-2x^{-\frac{5}{2}}\), its derivative \(y^\prime=-2\times(-\frac{5}{2})x^{-\frac{5}{2}-1}=5x^{-\frac{7}{2}}\).

Answer:

\(f^\prime(x)=27x^{\frac{7}{2}} + 5x^{-\frac{7}{2}}\)