Question

1. use logarithmic differentiation to find the derivative of the function y = (cos x)^x

Explanation:

Step1: Take natural logarithm on both sides

Given \(y = (\cos x)^x\), take \(\ln\) on both sides: \(\ln y=\ln(\cos x)^x\).
By the power - rule of logarithms \(\ln a^b = b\ln a\), we get \(\ln y=x\ln(\cos x)\).

Step2: Differentiate both sides with respect to \(x\)

Differentiate the left - hand side: Using the chain - rule, if \(u = y\), then \(\frac{d}{dx}(\ln y)=\frac{1}{y}\cdot y'\).
Differentiate the right - hand side: Using the product - rule \((uv)^\prime=u^\prime v + uv^\prime\), where \(u = x\) and \(v=\ln(\cos x)\).
We know that \(\frac{d}{dx}(x)=1\) and \(\frac{d}{dx}(\ln(\cos x))=\frac{1}{\cos x}\cdot(-\sin x)=-\tan x\) (by the chain - rule \(\frac{d}{dx}(\ln u)=\frac{u^\prime}{u}\) with \(u = \cos x\)).
So \(\frac{1}{y}\cdot y'=1\cdot\ln(\cos x)+x\cdot(-\tan x)\).

Step3: Solve for \(y'\)

Multiply both sides by \(y\). Since \(y = (\cos x)^x\), we have \(y'=(\cos x)^x(\ln(\cos x)-x\tan x)\).

Answer:

\(y'=(\cos x)^x(\ln(\cos x)-x\tan x)\)