Answer to Question #198871 in Calculus for Snakho

Question #198871

y(x) =(sinx) linx


1
Expert's answer
2021-05-27T06:04:28-0400
y=(sinx)lnxy=(\sin x)^{\ln x}

Then


lny=ln(sinx)lnx\ln y=\ln{(\sin x)^{\ln x}}

lny=lnxln(sinx)\ln y=\ln x\cdot\ln{(\sin x)}

Differentiate both sides with respect to xx


ddx(lny)=ddx(lnxln(sinx))\dfrac{d}{dx}(\ln y)=\dfrac{d}{dx}(\ln x\cdot\ln{(\sin x)})

Use the Chain Rule


1ydydx=ln(sinx)x+lnxsinxcosx\dfrac{1}{y}\cdot\dfrac{dy}{dx}=\dfrac{\ln (\sin x)}{x}+\dfrac{\ln x}{\sin x}\cdot \cos x

dydx=y(ln(sinx)x+lnxtanx)\dfrac{dy}{dx}=y(\dfrac{\ln (\sin x)}{x}+\dfrac{\ln x}{\tan x})

dydx=(sinx)lnx(ln(sinx)x+lnxtanx)\dfrac{dy}{dx}=(\sin x)^{\ln x}\bigg(\dfrac{\ln (\sin x)}{x}+\dfrac{\ln x}{\tan x}\bigg)



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