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()(cos(x)ln(cos((x)sin(x)) is what i got, although i am not entirely sure if this is correct.

If , then

Differentiating both sides gives

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

Step-by-step 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)]

do you mean add both of the equations together

because if you do the answer is 10x-5

step-by-step explanation: