`y = (tan(x))^(1/x)` Use logarithmic differentiation to find the derivative of the function.

Textbook Question

Chapter 3, 3.6 - Problem 49 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`y=(tanx)^(1/x)`

Taking natural logarithm of both the sides, and applying the properties of the logarithm

`lny=(1/x)lntanx`

Differentiating both sides with respect to x, we get

`1/ydy/dx=(1/x)d/dx(lntanx)+lntanx d/dx(x^-1)`

`1/ydy/dx=(1/x)(sec^2(x))/(tanx)+(-1x^(-2))lntanx`

`1/ydy/dx=(sec(x)csc(x))/(x)-lntan(x)/(x^2)`

`dy/dx=(tan(x))^(1/x)((sec(x)csc(x))/(x)-lntan(x)/(x^2))`

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hkj1385 | (Level 1) Assistant Educator

Posted on

`y = (tanx)^(1/x)`

 taking log to the base 'e' both sides we get

lny = (1/x)*tanx

differentiating both sides we get

(1/y)*dy/dx = `[(1/x)*sec^2x] + [-(1/x^2)*tanx]`

`or, dy/dx = y*[{(1/x)*sec^2x} - {(1/x^2)*tanx}]`

`or, dy/dx = {(tanx)^(1/x)}*[{(1/x)*sec^2x}-{(1/x^2)*tanx}]`

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