Question #226792

(Cos(2x))^1/(x^2) please help with the derivative. Thanks

Expert's answer

y=(cos⁡(2x))1/x2y=(\cos(2x))^{1/x^2}

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

Differentiate both sides with respect to xx


y′y=−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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