Question

differentiate
$y = \ln(x + 8)(4x - 7)(5 - x)$
$\frac{d}{dx}\ln((x + 8)(4x - 7)(5 - x)) = \square$

Explanation:

Step1: Use logarithm properties

By the property \(\ln(abc)=\ln a+\ln b+\ln c\), \(y = \ln(x + 8)+\ln(4x-7)+\ln(5 - x)\)

Step2: Differentiate term - by - term

The derivative of \(\ln u\) with respect to \(x\) is \(\frac{u^\prime}{u}\) (by the chain rule \((\ln u)^\prime=\frac{1}{u}\cdot u^\prime\)).
For \(u=x + 8\), \(u^\prime=1\), so the derivative of \(\ln(x + 8)\) is \(\frac{1}{x + 8}\).
For \(u = 4x-7\), \(u^\prime=4\), so the derivative of \(\ln(4x - 7)\) is \(\frac{4}{4x-7}\).
For \(u=5 - x\), \(u^\prime=-1\), so the derivative of \(\ln(5 - x)\) is \(\frac{-1}{5 - x}=\frac{1}{x - 5}\)

Answer:

\(\frac{1}{x + 8}+\frac{4}{4x-7}+\frac{1}{x - 5}\)