Find \( \frac{d}{dx}sinh^{-1}(x) \)

\begin{align} \end{align}

\begin{align} Let\quad y & = sinh^{-1}(x) \\ Then\quad x & = sinh(y) \\ And\quad \frac {dx}{dy}& = cosh(y) \\ & = +\sqrt{sinh^2(y) + 1 } \quad(+ve\ only,\ because\ cosh(y)\gt 0)\\ & = \sqrt{ (x^2 + 1)} \\ \therefore \quad \frac{dy}{dx} & = \frac{1}{\sqrt{x^2+1}} \end{align}

Find \( \frac{d}{dx}cosh^{-1}(x) \)

\begin{align} \end{align}

\begin{align} Let\quad y & = cosh^{-1}(x) \\ Then\quad x & = cosh(y) \\ And\quad \frac {dx}{dy}& = sinh(y) \\ & = +\sqrt{cosh^2(y) - 1 } \quad(+ve\ only,\ because\ sinh(y)\gt 0\ for\ these\ values\ of\ y.)\\ & = \sqrt{ x^2 - 1} \\ \therefore \quad \frac{dy}{dx} & = \frac{1}{\sqrt{x^2-1}} \end{align}

Find \( \frac{d}{dx}tanh^{-1}(x) \)

\begin{align} \end{align}

\begin{align} Let\quad y & = tanh^{-1}(x) \\ Then\quad x & = tanh(y) \\ And\quad \frac {dx}{dy}& = sech^2(y) \\ & = 1- tanh^2(y)\\ & = 1-x^2 \\ \therefore \quad \frac{dy}{dx} & = \frac{1}{1-x^2} \\ \end{align}