`f(x) = coth^-1(x^2)` Find the derivative of the function

`f(x) = coth^-1(x^2)`

The derivative formula of inverse hyperbolic cotangent is

• `d/dx[coth^(-1)(u)]=1/(1-u^2)*(du)/dx`

Applying this formula, the derivative of the function will be

`f'(x)=d/dx[coth^(-1) (x^2)]`

`f'(x)=1/(1-(x^2)^2) * d/dx(x^2)`

`f'(x)=1/(1-x^4) * d/dx(x^2)`

To take the derivative of `x^2` , apply the formula

• `d/dx(x^n)= n*x^(n-1)`

So, f'(x) will become

`f'(x)=1/(1-x^4)* 2x`

`f'(x)=(2x)/(1-x^4)`

Therefore, the derivative of the function is `f'(x)=(2x)/(1-x^4)` .