8. 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))