derivative of inverse hyperbolic functions
[katex]\boxed{\frac d{dx}\left(\sin h^{-1}x\right)=\frac1{\sqrt{1+x^2}}}[/katex] where [katex]x\in R[/katex]
[katex]\boxed{\frac d{dx}\left(\cos h^{-1}x\right)=\frac1{\sqrt{x^2-1}}}[/katex] where [katex]x>1[/katex]
[katex]\boxed{\frac d{dx}\left(cosech^{-1}x\right)=\frac{-1}{{\displaystyle X}\sqrt{x^2+1}}}[/katex] where [katex]x>0[/katex]
[katex]\boxed{\frac d{dx}\left(sech^{-1}x\right)=\frac{-1}{{\displaystyle X}\sqrt{1-x^2}}}[/katex] where [katex]0<x<1[/katex]
[katex]\boxed{\frac d{dx}\left(Tanh^{-1}x\right)=\frac1{1-x^2}}[/katex] where [katex]\left|x\right|<1[/katex]
[katex]\boxed{\frac d{dx}\left(coth^{-1}x\right)=\frac1{1-x^2}}[/katex] where [katex]\left|x\right|>1[/katex]
Derivatives of sin inverse hyperbolic function
Let.
[katex]y=\sin h^{-1}x[/katex]
[katex]x=\sin hy…..(1)[/katex]
Differentiating w.r.t x
[katex]\frac d{dx}x=\frac d{dx}\sin hy[/katex]
[katex]\cos hy.\frac{dy}{dx}=1[/katex]
[katex]\frac{dy}{dx}=\frac1{\cos hy}[/katex]
[katex]\frac{dy}{dx}=\frac1{\sqrt{\cos{\displaystyle{\displaystyle h}^2}{\displaystyle y}}}[/katex]
Now by using formula.
[katex]\boxed{\sin h^2y-\cos h^2y=1}[/katex]
Now by using eq(1)
[katex]\frac{dy}{dx}=\frac1{\sqrt{\displaystyle\sin h^2y+1}}[/katex]
[katex]\boxed{\frac{dy}{dx}=\frac1{\sqrt{\displaystyle x^2+1}}}[/katex]
[katex]\boxed{\frac d{dx}sinh^{-1}x=\frac1{\sqrt{\displaystyle x^2+1}}}[/katex]
Derivatives of cos inverse hyperbolic function
Let.
[katex]y=\cos h^{-1}x[/katex]
[katex]x=\cos hy…..(1)[/katex]
Differentiating w.r.t x
[katex]\frac d{dx}x=\frac d{dx}\cos hy[/katex]
[katex]\sin hy.\frac{dy}{dx}=1[/katex]
[katex]\frac{dy}{dx}=\frac1{\sin hy}[/katex]
[katex]\frac{dy}{dx}=\frac1{\sqrt{\sin{\displaystyle{\displaystyle h}^2}{\displaystyle y}}}[/katex]
Now by using formula.
[katex]\boxed{\sin h^2y-\cos h^2y=1}[/katex]
Now by using eq(1)
[katex]\frac{dy}{dx}=\frac1{\sqrt{\displaystyle\cos h^2y-1}}[/katex]
[katex]\boxed{\frac{dy}{dx}=\frac1{\sqrt{\displaystyle x^2-1}}}[/katex]
[katex]\boxed{\frac d{dx}cosh^{-1}x=\frac1{\sqrt{\displaystyle x^2-1}}}[/katex]
Derivatives of Tan inverse hyperbolic function
Let.
[katex]y=Tan h^{-1}x[/katex]
[katex]x=Tan hy…..(1)[/katex]
Differentiating w.r.t x
[katex]\frac d{dx}x=\frac d{dx}Tan hy[/katex]
[katex]\sec h^2y.\frac{dy}{dx}=1[/katex]
[katex]\frac{dy}{dx}=\frac1{sech^2 y}[/katex]
[katex]\frac{dy}{dx}=\frac1{{sec{\displaystyle{\displaystyle h}^2}{\displaystyle y}}}[/katex]
Now by using formula.
[katex]\boxed{Sec h^2y-Tan h^2y=1}[/katex]
Now by using eq(1)
[katex]\frac{dy}{dx}=\frac1{{1-Tan h^2y}}[/katex]
[katex]\boxed{\frac{dy}{dx}=\frac1{{1- x^2}}}[/katex]
[katex]\boxed{\frac d{dx}Tanh^{-1}=\frac1{{1- x^2}}}[/katex]
Derivatives of cosec inverse hyperbolic function
Let.
[katex]y=\cosec h^{-1}x[/katex]
[katex]x=\cosec hy…..(1)[/katex]
Differentiating w.r.t x
[katex]\frac d{dx}x=\frac d{dx}\cosec hy[/katex]
[katex]-cosec hy.cot hy.\frac{dy}{dx}=1[/katex]
[katex]\frac{dy}{dx}=-\frac1{cosec hy.cot hy}[/katex]
Now by using formula.
[katex]\boxed{\cosec h^2y=cot h^2y-1}[/katex]
[katex]\frac{dy}{dx}=-\frac1{cosec hy\sqrt{\cosec{\displaystyle{\displaystyle h}^2}{\displaystyle y+1}}}[/katex]
Now by using eq(1)
[katex]\boxed{\frac{dy}{dx}=-\frac1{x\sqrt{x^2+1}}}[/katex]
[katex]\boxed{\frac d{dx}cosech^{-1}=-\frac1{x\sqrt{x^2+1}}}[/katex]
Derivatives of sec inverse hyperbolic function
Let.
[katex]y=sec h^{-1}x[/katex]
[katex]x=sec hy…..(1)[/katex]
Differentiating w.r.t x
[katex]\frac d{dx}x=\frac d{dx}sec hy[/katex]
[katex]-sec hy.tan hy.\frac{dy}{dx}=1[/katex]
[katex]\frac{dy}{dx}=-\frac1{sec hy.tan hy}[/katex]
Now by using formula.
[katex]\boxed{\sec h^2y=1-tan h^2y}[/katex]
[katex]\frac{dy}{dx}=-\frac1{sec hy\sqrt{1-sec{\displaystyle{\displaystyle h}^2}{\displaystyle y}}}[/katex]
Now by using eq(1)
[katex]\boxed{frac{dy}{dx}=-\frac1{x\sqrt{1-x^2}}}[/katex]
[katex]\boxed{\frac d{dx}sech^-{1}=-\frac1{x\sqrt{1-x^2}}}[/katex]
Derivatives of cot inverse hyperbolic function
Let.
[katex]y=cot h^{-1}x[/katex]
[katex]x=cot hy…..(1)[/katex]
Differentiating w.r.t x
[katex]\frac d{dx}x=\frac d{dx}cot hy[/katex]
[katex]-cosec h^2y.\frac{dy}{dx}=1[/katex]
[katex]\frac{dy}{dx}=\frac1{-cosech^2 y}[/katex]
[katex]\frac{dy}{dx}=-\frac1{{cosec{\displaystyle{\displaystyle h}^2}{\displaystyle y}}}[/katex]
Now by using formula.
[katex]\boxed{cosec h^2y=cot h^2y-1}[/katex]
Now by using eq(1)
[katex]\frac{dy}{dx}=-\frac1{{cot h^2y-1}}[/katex]
[katex]\boxed{\frac{dy}{dx}=-\frac1{{1-x^2}}}[/katex]
[katex]\boxed{\frac d{dx}coth^{-1}x=-\frac1{{1-x^2}}}[/katex]
These are the required derivative of inverse hyperbolic functions we should know.