Question

find the derivative of $f(x)=6\sqrt{x}+\frac{8}{x^{3}}$.
type your answer without fractional or negative exponents.
$f(x)=\square$
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Explanation:

Step1: Rewrite the function

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

Step2: Apply the power rule

The power rule is \((x^{n})^\prime=nx^{n - 1}\).
For the first term: \((6x^{\frac{1}{2}})^\prime=6\times\frac{1}{2}x^{\frac{1}{2}-1}=3x^{-\frac{1}{2}}\).
For the second term: \((8x^{-3})^\prime=8\times(-3)x^{-3 - 1}=-24x^{-4}\).

Step3: Rewrite without negative exponents

\(3x^{-\frac{1}{2}}=\frac{3}{\sqrt{x}}\) and \( - 24x^{-4}=-\frac{24}{x^{4}}\).

Answer:

\(\frac{3}{\sqrt{x}}-\frac{24}{x^{4}}\)