Double Math

# Derivative of Inverse Hyperbolic Functions

derivative of inverse hyperbolic functions

\boxed{\frac d{dx}\left(\sin h^{-1}x\right)=\frac1{\sqrt{1+x^2}}} where x\in R

\boxed{\frac d{dx}\left(\cos h^{-1}x\right)=\frac1{\sqrt{x^2-1}}} where x>1

\boxed{\frac d{dx}\left(cosech^{-1}x\right)=\frac{-1}{{\displaystyle X}\sqrt{x^2+1}}} where x>0

\boxed{\frac d{dx}\left(sech^{-1}x\right)=\frac{-1}{{\displaystyle X}\sqrt{1-x^2}}} where 0<x<1

\boxed{\frac d{dx}\left(Tanh^{-1}x\right)=\frac1{1-x^2}} where \left|x\right|<1

\boxed{\frac d{dx}\left(coth^{-1}x\right)=\frac1{1-x^2}} where \left|x\right|>1

## Let.

y=\sin h^{-1}x

x=\sin hy.....(1)

Differentiating w.r.t x

\frac d{dx}x=\frac d{dx}\sin hy

\cos hy.\frac{dy}{dx}=1

\frac{dy}{dx}=\frac1{\cos hy}

\frac{dy}{dx}=\frac1{\sqrt{\cos{\displaystyle{\displaystyle h}^2}{\displaystyle y}}}

Now by using formula.

\boxed{\sin h^2y-\cos h^2y=1}

Now by using eq(1)

\frac{dy}{dx}=\frac1{\sqrt{\displaystyle\sin h^2y+1}}

\boxed{\frac{dy}{dx}=\frac1{\sqrt{\displaystyle x^2+1}}}

\boxed{\frac d{dx}sinh^{-1}x=\frac1{\sqrt{\displaystyle x^2+1}}}

## Let.

y=\cos h^{-1}x

x=\cos hy.....(1)

Differentiating w.r.t x

\frac d{dx}x=\frac d{dx}\cos hy

\sin hy.\frac{dy}{dx}=1

\frac{dy}{dx}=\frac1{\sin hy}

\frac{dy}{dx}=\frac1{\sqrt{\sin{\displaystyle{\displaystyle h}^2}{\displaystyle y}}}

Now by using formula.

\boxed{\sin h^2y-\cos h^2y=1}

Now by using eq(1)

\frac{dy}{dx}=\frac1{\sqrt{\displaystyle\cos h^2y-1}}

\boxed{\frac{dy}{dx}=\frac1{\sqrt{\displaystyle x^2-1}}}

\boxed{\frac d{dx}cosh^{-1}x=\frac1{\sqrt{\displaystyle x^2-1}}}

## Let.

y=Tan h^{-1}x

x=Tan hy.....(1)

Differentiating w.r.t x

\frac d{dx}x=\frac d{dx}Tan hy

\sec h^2y.\frac{dy}{dx}=1

\frac{dy}{dx}=\frac1{sech^2 y}

\frac{dy}{dx}=\frac1{{sec{\displaystyle{\displaystyle h}^2}{\displaystyle y}}}

Now by using formula.

\boxed{Sec h^2y-Tan h^2y=1}

Now by using eq(1)

\frac{dy}{dx}=\frac1{{1-Tan h^2y}}

\boxed{\frac{dy}{dx}=\frac1{{1- x^2}}}

\boxed{\frac d{dx}Tanh^{-1}=\frac1{{1- x^2}}}

## Let.

y=\cosec h^{-1}x

x=\cosec hy.....(1)

Differentiating w.r.t x

\frac d{dx}x=\frac d{dx}\cosec hy

-cosec hy.cot hy.\frac{dy}{dx}=1

\frac{dy}{dx}=-\frac1{cosec hy.cot hy}

Now by using formula.

\boxed{\cosec h^2y=cot h^2y-1}

\frac{dy}{dx}=-\frac1{cosec hy\sqrt{\cosec{\displaystyle{\displaystyle h}^2}{\displaystyle y+1}}}

Now by using eq(1)

\boxed{\frac{dy}{dx}=-\frac1{x\sqrt{x^2+1}}}

\boxed{\frac d{dx}cosech^{-1}=-\frac1{x\sqrt{x^2+1}}}

## Let.

y=sec h^{-1}x

x=sec hy.....(1)

Differentiating w.r.t x

\frac d{dx}x=\frac d{dx}sec hy

-sec hy.tan hy.\frac{dy}{dx}=1

\frac{dy}{dx}=-\frac1{sec hy.tan hy}

Now by using formula.

\boxed{\sec h^2y=1-tan h^2y}

\frac{dy}{dx}=-\frac1{sec hy\sqrt{1-sec{\displaystyle{\displaystyle h}^2}{\displaystyle y}}}

Now by using eq(1)

\boxed{frac{dy}{dx}=-\frac1{x\sqrt{1-x^2}}}

\boxed{\frac d{dx}sech^-{1}=-\frac1{x\sqrt{1-x^2}}}

## Let.

y=cot h^{-1}x

x=cot hy.....(1)

Differentiating w.r.t x

\frac d{dx}x=\frac d{dx}cot hy

-cosec h^2y.\frac{dy}{dx}=1

\frac{dy}{dx}=\frac1{-cosech^2 y}

\frac{dy}{dx}=-\frac1{{cosec{\displaystyle{\displaystyle h}^2}{\displaystyle y}}}

Now by using formula.

\boxed{cosec h^2y=cot h^2y-1}

Now by using eq(1)

\frac{dy}{dx}=-\frac1{{cot h^2y-1}}

\boxed{\frac{dy}{dx}=-\frac1{{1-x^2}}}

\boxed{\frac d{dx}coth^{-1}x=-\frac1{{1-x^2}}}

These are the required derivative of inverse hyperbolic functions we should know.