Question

hw 14 - derivatives of logs section 2.9: problem 5 (1 point)
compute the derivatives of the given functions.
a) $f(x)=ln(7x^{8})$, $f(x)=square$.
b) $g(x)=ln(sqrt3{x})$, $g(x)=square$.
note: you can earn partial credit on this problem.
preview my answers check answers
you have attempted this problem 0 times.
this homework set is closed.
email instructor

Explanation:

Step1: Use log - property for \(f(x)\)

First, use the property \(\ln(ab)=\ln a+\ln b\). So \(f(x)=\ln(7x^{8})=\ln7 + \ln(x^{8})\). Since \(\ln7\) is a constant, its derivative is 0. And by the power - rule for logarithms \(\ln(x^{n})=n\ln x\), we have \(\ln(x^{8}) = 8\ln x\). The derivative of \(\ln x\) is \(\frac{1}{x}\), so \(f^{\prime}(x)=0 + 8\times\frac{1}{x}=\frac{8}{x}\).

Step2: Use log - property for \(g(x)\)

Rewrite \(g(x)=\ln(\sqrt[3]{x})=\ln(x^{\frac{1}{3}})\). By the power - rule for logarithms \(\ln(x^{n})=n\ln x\), we get \(g(x)=\frac{1}{3}\ln x\). The derivative of \(\ln x\) is \(\frac{1}{x}\), so \(g^{\prime}(x)=\frac{1}{3}\times\frac{1}{x}=\frac{1}{3x}\).

Answer:

a) \(\frac{8}{x}\)
b) \(\frac{1}{3x}\)