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

derivative of inverse hyperbolic functions

\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

Derivatives of sin inverse hyperbolic function

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}}}

Derivatives of cos inverse hyperbolic function

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}}}

Derivatives of Tan inverse hyperbolic function

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}}}

Derivatives of cosec inverse hyperbolic function

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}}}

Derivatives of sec inverse hyperbolic function

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}}}

Derivatives of cot inverse hyperbolic function

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.

Spread the love
Azhar Ali

Azhar Ali

I graduated in Mathematics from the University of Sargodha, having master degree in Mathematics.

Leave a Reply

Your email address will not be published.

Mathematics is generally known as Math in US and Maths in the UK.

Contact Us

Copyright by Double Math. All Right Reserved 2019 to 2022