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.

Step-by-step explanation:

y = (cos x)^(sin x)

Take log both sides.

ln y = ln ((cos x)^(sin x))

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

Take derivative.

1/y dy/dx = (sin x) (1/cos x) (-sin x) + (cos x) ln (cos x)

dy/dx = y [ (cos x) ln (cos x) − sin x tan x ]

dy/dx = (cos x)^(sin x) [ (cos x) ln (cos x) − sin x tan x ]

Your answer is correct.

[(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)]

the figures below are made out of circles, semicircles, quarter … find the area and the perimeter of each figure and give your answers as a completely simplified exact value in terms of π (no approximations). remember that we'll also have to cut whatever areas we get in half, since we're only dealing

step-by-step explanation:

this is a very good question

step-by-step explanation:

i got a hair cut today from my mom not to flex just to flex kim jon un is one rad fellow i'd say you'd need to take the hairline in kinda like a box shape like someone rammed a lego brick into your head. from there you really need to take the top up keeping with that boxy aesthetic, finally coat the whole thing with gel if your country economics are failing or whatever you can just coat it with the starving children's food rations. about 5 orphans per strand is a good ratio. good luck with the fresh cut my dude