`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)` .
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