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

Sunday, December 7, 2014

$\displaystyle{f{{\left({x}\right)}}}={{\coth}}^{{-{{1}}}}{\left({{x}}^{{2}}\right)}$

The derivative formula of inverse hyperbolic cotangent is

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left[{{\coth}}^{{-{1}}}{\left({u}\right)}\right]}=\frac{{1}}{{{1}-{{u}}^{{2}}}}\cdot\frac{{{d}{u}}}{{\left.{d}{x}}}$

Applying this formula, the derivative of the function will be

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{d}}{{\left.{d}{x}}}{\left[{{\coth}}^{{-{1}}}{\left({{x}}^{{2}}\right)}\right]}$

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{1}}{{{1}-{{\left({{x}}^{{2}}\right)}}^{{2}}}}\cdot\frac{{d}}{{\left.{d}{x}}}{\left({{x}}^{{2}}\right)}$

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{1}}{{{1}-{{x}}^{{4}}}}\cdot\frac{{d}}{{\left.{d}{x}}}{\left({{x}}^{{2}}\right)}$

To take the derivative of $\displaystyle{{x}}^{{2}}$ , apply the formula

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left({{x}}^{{n}}\right)}={n}\cdot{{x}}^{{{n}-{1}}}$

So, f'(x) will become

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{1}}{{{1}-{{x}}^{{4}}}}\cdot{2}{x}$

$\displaystyle{f{'}}{\left({x}\right)}=\frac{{{2}{x}}}{{{1}-{{x}}^{{4}}}}$

Therefore, the derivative of the function is $\displaystyle{f{'}}{\left({x}\right)}=\frac{{{2}{x}}}{{{1}-{{x}}^{{4}}}}$ .

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