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