Question

Use logarithmic differentiation to find dy/dx if y=(cosx)sinx. Write the derivative in terms of x only. Show all your work.

(cos ^sin(x))(x)(cos(x)ln(cos(x))-(tan(x)sin(x)) is what I got, although I am not entirely sure if this is correct.

Answer 1

[(cosx)^(sinx)][-sinx(tanx) + (cosx)ln(cosx)]

Explanation:

y = (cosx)^(sinx)

lny = (sinx) × ln(cosx)

(1/y)×dy/dx = (sinx)(1/cosx)(-sinx) + (cosx)ln(cosx)

(1/y)dy/dx = -sinxtanx + (cosx)ln(cosx)

dy/dx = y[-sinxtanx + (cosx)ln(cosx)]

dy/dx =

[(cosx)^(sinx)][-sinxtanx + (cosx)ln(cosx)]

Answer 2

If
`y=(\cos x)^(\sin x)`, then

`\ln y=\sin x\ln(\cos x)`

Differentiating both sides gives

`\frac1y(\mathrm dy)/(\mathrm dx)=\cos x\ln(\cos x)+\sin x\left(-(\sin x)/(\cos x)\right)=\cos x\ln(\cos x)-\sin x\tan x`

`\implies(\mathrm dy)/(\mathrm dx)=(\cos x)^(\sin x)\left(\cos x\ln(\cos x)-\sin x\tan x\right)`