Derivative of Inverse Hyperbolic Functions

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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.

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