Valishati
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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=

Respuesta :

LammettHash

If [tex]y=(\cos x)^{\sin x}[/tex], then

[tex]\ln y=\sin x\ln(\cos x)[/tex]

Differentiating both sides gives

[tex]\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}=\cos x\ln(\cos x)+\sin x\left(-\dfrac{\sin x}{\cos x}\right)=\cos x\ln(\cos x)-\sin x\tan x[/tex]

[tex]\implies\dfrac{\mathrm dy}{\mathrm dx}=(\cos x)^{\sin x}\left(\cos x\ln(\cos x)-\sin x\tan x\right)[/tex]

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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