This page breaks down the derivatives of inverse trigonometric functions such as arcsin, arccos, arctan, arccot, arccsc, and arcsec. You’ll find a formula reference sheet, and many practice problems with answers to help you master this essential calculus skill.
Notes

Practice Problems
Find the derivative of each
\(\textbf{1)}\) \(f(x)=\cos^2(x)+3\sin^{−1}(x), \text{find } f'(x)\)
The derivative is \(f'(x)=-2\sin(x)\cos(x)+\displaystyle \frac{3}{\sqrt{1-x^2}}\)
\(\textbf{2)}\) \(f(x)=8\sin^{−1}(x)−2\csc^{−1}(x), \text{find } f'(x)\)
The derivative is \(f'(x)=\displaystyle \frac{8}{\sqrt{1-x^2}}+\frac{2}{|x|\sqrt{x^2-1}}\)
\(\textbf{3)}\) \(f(x)=\arctan(2x)+4\tan(3x), \text{find } f'(x)\)
The derivative is \(f'(x)=\displaystyle\frac{2}{4x^2+1}+12\sec ^2\left(3x\right)
\)
\(\textbf{4)}\) \(f(x)=\sec^{−1}(x)−\cos^{−1}(x), \text{find } f'(x)\)
The derivative is \(f'(x) =\displaystyle\frac{1}{|x|\sqrt{x^2-1}}+\frac{1}{\sqrt{1-x^2}}\)
\(\textbf{5)}\) \(f(x)=x^3\arcsin(2x), \text{find } f'(x)\)
The derivative is \(f'(x) =3x^2\arcsin \left(2x\right)+\displaystyle\frac{2x^3}{\sqrt{1-4x^2}}\)
\(\textbf{6)}\) \(f(x)=\displaystyle \frac{\sin^{−1}(x)}{x}, \text{find } f'(x)\)
The derivative is \(f'(x) =\displaystyle\frac{\frac{x}{\sqrt{1-x^2}}-\sin^{-1} \left(x\right)}{x^2}=\frac{x-\sin^{-1} \left(x\right)\sqrt{1-x^2}}{x^2\sqrt{1-x^2}}\)
\(\textbf{7)}\) \(f(x)=3x^5+\tan^{−1}(3x^5), \text{find } f'(x)\)
The derivative is \(f'(x) =15x^4+\displaystyle\frac{15x^4}{9x^{10}+1}\,\,= \,\,\displaystyle \frac{135x^{14}+30x^4}{9x^{10}+1}\)
\(\textbf{8)}\) \(f(x)=\sin^{−1}(\ln x), \text{find } f'(x)\)
The derivative is \(f'(x) =\displaystyle \frac{1}{x\sqrt{1-\left(\ln x \right)^2}}\)
\(\,\,\,\,\,f(x)=\sin^{−1}(\ln x)\)
\(\,\,\,\,\,u=\ln x, u’=\frac{1}{x}\)
\(\,\,\,\,\,\frac{d}{dx} \arcsin{u}=\displaystyle \frac{u’}{\sqrt{1-u^2}}\)
\(\,\,\,\,\,\frac{d}{dx} \arcsin{u}=\displaystyle \frac{\frac{1}{x}}{\sqrt{1-(\ln{x})^2}}\)
\(\,\,\,\,\,f'(x) =\displaystyle \frac{1}{x\sqrt{1-\left(\ln x \right)^2}}\)
\(\textbf{9)}\) \(f(x)=\ln (\arcsin(x)), \text{find } f'(x)\)
The derivative is \(f'(x) =\displaystyle \frac{1}{\arcsin \left(x\right)\sqrt{1-x^2}}\)
\(\textbf{10)}\) \(f(x) = \cot^{-1}(2x), \text{find } f'(x)\)
The derivative is \(f'(x) =-\displaystyle \frac{2}{4x^2+1}\)
\(\textbf{11)}\) \(y = \tan^{-1}(x^2+4x), \text{find } \displaystyle\frac{dy}{dx}\)
The derivative is \(\displaystyle\frac{dy}{dx} =\displaystyle \frac{2x+4}{1+\left(x^2+4x\right)^2}\)
See Related Pages\(\)
In Summary
Inverse trigonometric functions are first introduced to solve problems involving unknown angles but known sides in right triangles. These functions include the inverse sine (arcsin), inverse cosine (arccos), and inverse tangent (arctan). Most people tend to memorize the inverse trig derivatives.
\(\displaystyle \frac{d}{dx} \arcsin(x)=\displaystyle \frac{1}{\sqrt{1-x^2}} \,\,\,\,\, x \ne \pm1 \)
\(\displaystyle \frac{d}{dx} \arccos(x)=\frac{-1}{\sqrt{1-x^2}} \,\,\,\,\, x \ne \pm1 \)
\(\displaystyle \frac{d}{dx} \arctan(x)=\frac{1}{1+x^2} \)
\(\displaystyle \frac{d}{dx} \text{arccot}(x)=\frac{-1}{1+x^2} \)
\(\displaystyle \frac{d}{dx} \text{arcsec}(x)= \frac{1}{|x|\sqrt{x^2-1}} \,\,\,\,\, x \ne \pm1 \)
\(\displaystyle \frac{d}{dx} \text{arccsc}(x)=\frac{-1}{|x|\sqrt{x^2-1}} \,\,\,\,\, x \ne \pm1 \)