Question

exercsies 3.3 differentiation rules
score: 7/19 answered: 7/19
question 5
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let $f(x)=7 x^{6} sqrt{x}+\frac{-3}{x^{2} sqrt{x}}$.
$f^{prime}(x)=$
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Explanation:

Step1: Simplify the function

Use the rule \(a^m\times a^n=a^{m + n}\) and \(\frac{1}{a^m}=a^{-m}\).
For \(7x^{6}\sqrt{x}\), since \(\sqrt{x}=x^{\frac{1}{2}}\), then \(7x^{6}\sqrt{x}=7x^{6+\frac{1}{2}} = 7x^{\frac{13}{2}}\).
For \(\frac{-3}{x^{2}\sqrt{x}}\), since \(\sqrt{x}=x^{\frac{1}{2}}\), then \(\frac{-3}{x^{2}\sqrt{x}}=-3x^{-2-\frac{1}{2}}=-3x^{-\frac{5}{2}}\). So \(f(x)=7x^{\frac{13}{2}}-3x^{-\frac{5}{2}}\).

Step2: Apply the power rule

The power rule is \((x^n)^\prime=nx^{n - 1}\).
For \(y = 7x^{\frac{13}{2}}\), \(y^\prime=7\times\frac{13}{2}x^{\frac{13}{2}-1}=\frac{91}{2}x^{\frac{11}{2}}\).
For \(y=-3x^{-\frac{5}{2}}\), \(y^\prime=-3\times(-\frac{5}{2})x^{-\frac{5}{2}-1}=\frac{15}{2}x^{-\frac{7}{2}}\).

Answer:

\(f^\prime(x)=\frac{91}{2}x^{\frac{11}{2}}+\frac{15}{2x^{\frac{7}{2}}}\)