Question

Differentiate (sin/x)^1/x^2 ?

Answer 1

`dy/dx=(sinx)^(1/x^2)[(1/x^2)cot(x)-(2/x^3)(log(sinx))]`

Explanation:

Let `y=(sinx)^(1/x^2)`
taking ln both side we get.
`lny=ln((sinx)^(1/x^2))`
apply `ln(a^m)=mlna` we get
`lny=(1/x^2)(ln(sinx))`
differentiate both side wrt x we get
`(1/y)(dy/dx)=(1/x^2)(cosx/sinx)+(ln(sinx))(-2/x^3)`
`(1/y)(dy/dx)=(1/x^2)cot(x)-(2/x^3)(ln(sinx))`
`dy/dx=y[(1/x^2)cot(x)-(2/x^3)(ln(sinx))]`
putting `y=(sinx)^(1/x^2)` in above equation we get
`dy/dx=(sinx)^(1/x^2)[(1/x^2)cot(x)-(2/x^3)(ln(sinx))]`
this is the required result
hope you understand

Answer 2

`(sinx)^(1/x^2)*(cosx/(x^2sinx)-(2ln(sinx))/x^3)`

Explanation:

We have: `(sinx)^(1/x^2)`

We rewrite this as:

`(sinx)^(x^-2)`

Remember this rule:

`d/dx[f(x)^(g(x))]=f(x)^(g(x))*d/dx[ln(f(x))*g(x)]`

Also, remember the following:

`d/dx[lnx]=1/x`

`d/dx[f(x)*g(x)]=f'(x)*g(x)+f(x)*g'(x)`

`d/dx[f(g(x))]=f'(g(x))*g'(x)`

We now have:

`(sinx)^(x^-2)*d/dx[ln(sinx)*x^-2]`

`(sinx)^(x^-2)*(d/dx[ln(sinx)]*x^-2+ln(sinx)*d/dx[x^-2])`

Some other rules:

`d/dx[x^n]=nx^(n-1)`

`d/dx[sinx]=cosx`

`=>(sinx)^(x^-2)*(1/(sinx)*x^-2*d/dx[sinx]+ln(sinx)(-2x^-3))`

`=>(sinx)^(x^-2)*(1/(sinx)*x^-2*cosx+ln(sinx)(-2x^-3))`

`=>(sinx)^(1/x^2)*(1/(sinx)*cosx*1/x^2+ln(sinx)(-2x^-3))`

`=>(sinx)^(1/x^2)*(cosx/((sinx)x^2)+ln(sinx)(-2x^-3))`

`=>(sinx)^(1/x^2)*(cosx/((sinx)x^2)-2ln(sinx)(1/x^3))`

`=>(sinx)^(1/x^2)*(cosx/(x^2sinx)-(2ln(sinx))/x^3)`

This is our answer!