Question #226792
(Cos(2x))^1/(x^2) please help with the derivative. Thanks
1
Expert's answer
2021-08-18T14:20:39-0400
y=(cos(2x))1/x2y=(\cos(2x))^{1/x^2}

lny=ln(cos(2x))x2\ln y=\dfrac{\ln(\cos(2x))}{x^2}

Differentiate both sides with respect to xx


yy=2x2tan(2x)2xln(cos(2x))x4\dfrac{y'}{y}=\dfrac{-2x^2\tan(2x)-2x\ln(\cos(2x))}{x^4}

y=2x3(xtan(2x)+ln(cos(2x)))(cos(2x))1/x2y'=-\dfrac{2}{x^3}\big(x\tan(2x)+\ln(\cos(2x))\big)\big(\cos(2x)\big)^{1/x^2}


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Comments

Amorah Chauke
19.08.21, 12:04

Thank you

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