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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.

Use logarithmic differentiation to find dydx if ycosxsinx Write the derivative in terms of x only Show all your work cos sinxxcosxlncosxtanxsinx is what I got a class=

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MathPhys

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.

amna04352

Answer:

[(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)]

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