`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))`

`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}]`

``

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now