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. These problems also include chain rule, product rule, and quotient rule examples involving inverse trig functions.
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 \(f'(x)=-2\sin(x)\cos(x)+\displaystyle \frac{3}{\sqrt{1-x^2}}\)
\(\,\,\,\,\,f(x)=\cos^2(x)+3\sin^{-1}(x)\)
\(\,\,\,\,\,\frac{d}{dx}\left[\cos^2(x)\right]=2\cos(x)(-\sin(x))\)
\(\,\,\,\,\,\frac{d}{dx}\left[3\sin^{-1}(x)\right]=3\cdot\frac{1}{\sqrt{1-x^2}}\)
\(\,\,\,\,\,f'(x)=-2\sin(x)\cos(x)+\displaystyle \frac{3}{\sqrt{1-x^2}}\)
\(\,\,\,\,\,\)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 \(f'(x)=\displaystyle \frac{8}{\sqrt{1-x^2}}+\frac{2}{|x|\sqrt{x^2-1}}\)
\(\,\,\,\,\,f(x)=8\sin^{-1}(x)-2\csc^{-1}(x)\)
\(\,\,\,\,\,\frac{d}{dx}\sin^{-1}(x)=\displaystyle\frac{1}{\sqrt{1-x^2}}\)
\(\,\,\,\,\,\frac{d}{dx}\csc^{-1}(x)=\displaystyle-\frac{1}{|x|\sqrt{x^2-1}}\)
\(\,\,\,\,\,f'(x)=8\left(\frac{1}{\sqrt{1-x^2}}\right)-2\left(-\frac{1}{|x|\sqrt{x^2-1}}\right)\)
\(\,\,\,\,\,f'(x)=\displaystyle \frac{8}{\sqrt{1-x^2}}+\frac{2}{|x|\sqrt{x^2-1}}\)
\(\,\,\,\,\,\)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 \(f'(x)=\displaystyle\frac{2}{4x^2+1}+12\sec ^2\left(3x\right)
\)
\(\,\,\,\,\,f(x)=\arctan(2x)+4\tan(3x)\)
\(\,\,\,\,\,\frac{d}{dx}\arctan(u)=\displaystyle\frac{u’}{1+u^2}\)
\(\,\,\,\,\,u=2x,\quad u’=2\)
\(\,\,\,\,\,\frac{d}{dx}\arctan(2x)=\displaystyle\frac{2}{1+(2x)^2}\)
\(\,\,\,\,\,\frac{d}{dx}\left[4\tan(3x)\right]=4\cdot3\sec^2(3x)\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{2}{4x^2+1}+12\sec ^2\left(3x\right)\)
\(\,\,\,\,\,\)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 \(f'(x) =\displaystyle\frac{1}{|x|\sqrt{x^2-1}}+\frac{1}{\sqrt{1-x^2}}\)
\(\,\,\,\,\,f(x)=\sec^{-1}(x)-\cos^{-1}(x)\)
\(\,\,\,\,\,\frac{d}{dx}\sec^{-1}(x)=\displaystyle\frac{1}{|x|\sqrt{x^2-1}}\)
\(\,\,\,\,\,\frac{d}{dx}\cos^{-1}(x)=\displaystyle-\frac{1}{\sqrt{1-x^2}}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{1}{|x|\sqrt{x^2-1}}-\left(-\frac{1}{\sqrt{1-x^2}}\right)\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{1}{|x|\sqrt{x^2-1}}+\frac{1}{\sqrt{1-x^2}}\)
\(\,\,\,\,\,\)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 \(f'(x) =3x^2\arcsin \left(2x\right)+\displaystyle\frac{2x^3}{\sqrt{1-4x^2}}\)
\(\,\,\,\,\,f(x)=x^3\arcsin(2x)\)
\(\,\,\,\,\,\text{Use the product rule.}\)
\(\,\,\,\,\,\frac{d}{dx}\left[x^3\right]=3x^2\)
\(\,\,\,\,\,\frac{d}{dx}\arcsin(2x)=\displaystyle\frac{2}{\sqrt{1-(2x)^2}}\)
\(\,\,\,\,\,f'(x)=3x^2\arcsin(2x)+x^3\left(\frac{2}{\sqrt{1-4x^2}}\right)\)
\(\,\,\,\,\,f'(x)=3x^2\arcsin \left(2x\right)+\displaystyle\frac{2x^3}{\sqrt{1-4x^2}}\)
\(\,\,\,\,\,\)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 \(f'(x) =\displaystyle\frac{\frac{x}{\sqrt{1-x^2}}-\sin^{-1} \left(x\right)}{x^2}=\frac{x-\sin^{-1} \left(x\right)\sqrt{1-x^2}}{x^2\sqrt{1-x^2}}\)
\(\,\,\,\,\,f(x)=\displaystyle \frac{\sin^{-1}(x)}{x}\)
\(\,\,\,\,\,\text{Use the quotient rule.}\)
\(\,\,\,\,\,\frac{d}{dx}\sin^{-1}(x)=\displaystyle\frac{1}{\sqrt{1-x^2}}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{x\left(\frac{1}{\sqrt{1-x^2}}\right)-\sin^{-1}(x)(1)}{x^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{\frac{x}{\sqrt{1-x^2}}-\sin^{-1}(x)}{x^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{x-\sin^{-1}(x)\sqrt{1-x^2}}{x^2\sqrt{1-x^2}}\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=\displaystyle\frac{\frac{x}{\sqrt{1-x^2}}-\sin^{-1}(x)}{x^2}\)
\(\textbf{7)}\) \(f(x)=3x^5+\tan^{−1}(3x^5), \text{find } f'(x)\)
The derivative is \(f'(x) =15x^4+\displaystyle\frac{15x^4}{9x^{10}+1}\,\,= \,\,\displaystyle \frac{135x^{14}+30x^4}{9x^{10}+1}\)
\(\,\,\,\,\,f(x)=3x^5+\tan^{-1}(3x^5)\)
\(\,\,\,\,\,\frac{d}{dx}\left[3x^5\right]=15x^4\)
\(\,\,\,\,\,u=3x^5,\quad u’=15x^4\)
\(\,\,\,\,\,\frac{d}{dx}\tan^{-1}(u)=\displaystyle\frac{u’}{1+u^2}\)
\(\,\,\,\,\,\frac{d}{dx}\tan^{-1}(3x^5)=\displaystyle\frac{15x^4}{1+(3x^5)^2}\)
\(\,\,\,\,\,f'(x)=15x^4+\displaystyle\frac{15x^4}{9x^{10}+1}\)
\(\,\,\,\,\,f'(x)=\displaystyle \frac{135x^{14}+30x^4}{9x^{10}+1}\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=15x^4+\displaystyle\frac{15x^4}{9x^{10}+1}\)
\(\textbf{8)}\) \(f(x)=\sin^{−1}(\ln x), \text{find } f'(x)\)
The derivative is \(f'(x) =\displaystyle \frac{1}{x\sqrt{1-\left(\ln x \right)^2}}\)
\(\,\,\,\,\,f(x)=\sin^{-1}(\ln x)\)
\(\,\,\,\,\,u=\ln x,\quad u’=\frac{1}{x}\)
\(\,\,\,\,\,\frac{d}{dx} \arcsin{u}=\displaystyle \frac{u’}{\sqrt{1-u^2}}\)
\(\,\,\,\,\,\frac{d}{dx} \arcsin{u}=\displaystyle \frac{\frac{1}{x}}{\sqrt{1-(\ln{x})^2}}\)
\(\,\,\,\,\,f'(x) =\displaystyle \frac{1}{x\sqrt{1-\left(\ln x \right)^2}}\)
\(\,\,\,\,\,\)The derivative is \(f'(x) =\displaystyle \frac{1}{x\sqrt{1-\left(\ln x \right)^2}}\)
\(\textbf{9)}\) \(f(x)=\ln (\arcsin(x)), \text{find } f'(x)\)
The derivative is \(f'(x) =\displaystyle \frac{1}{\arcsin \left(x\right)\sqrt{1-x^2}}\)
\(\,\,\,\,\,f(x)=\ln(\arcsin(x))\)
\(\,\,\,\,\,u=\arcsin(x)\)
\(\,\,\,\,\,\frac{d}{dx}\ln(u)=\displaystyle\frac{u’}{u}\)
\(\,\,\,\,\,u’=\displaystyle\frac{1}{\sqrt{1-x^2}}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{\frac{1}{\sqrt{1-x^2}}}{\arcsin(x)}\)
\(\,\,\,\,\,f'(x)=\displaystyle \frac{1}{\arcsin \left(x\right)\sqrt{1-x^2}}\)
\(\,\,\,\,\,\)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 \(f'(x) =-\displaystyle \frac{2}{4x^2+1}\)
\(\,\,\,\,\,f(x)=\cot^{-1}(2x)\)
\(\,\,\,\,\,u=2x,\quad u’=2\)
\(\,\,\,\,\,\frac{d}{dx}\cot^{-1}(u)=\displaystyle-\frac{u’}{1+u^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle-\frac{2}{1+(2x)^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle-\frac{2}{4x^2+1}\)
\(\,\,\,\,\,\)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 \(\displaystyle\frac{dy}{dx} =\displaystyle \frac{2x+4}{1+\left(x^2+4x\right)^2}\)
\(\,\,\,\,\,y=\tan^{-1}(x^2+4x)\)
\(\,\,\,\,\,u=x^2+4x,\quad u’=2x+4\)
\(\,\,\,\,\,\frac{d}{dx}\tan^{-1}(u)=\displaystyle\frac{u’}{1+u^2}\)
\(\,\,\,\,\,\displaystyle\frac{dy}{dx}=\displaystyle\frac{2x+4}{1+\left(x^2+4x\right)^2}\)
\(\,\,\,\,\,\)The derivative is \(\displaystyle\frac{dy}{dx}=\displaystyle\frac{2x+4}{1+\left(x^2+4x\right)^2}\)
\(\textbf{12)}\) \(f(x)=\arccos(5x), \text{find } f'(x)\)
The derivative is \(f'(x)=\displaystyle-\frac{5}{\sqrt{1-25x^2}}\)
\(\,\,\,\,\,f(x)=\arccos(5x)\)
\(\,\,\,\,\,u=5x,\quad u’=5\)
\(\,\,\,\,\,\frac{d}{dx}\arccos(u)=\displaystyle-\frac{u’}{\sqrt{1-u^2}}\)
\(\,\,\,\,\,f'(x)=\displaystyle-\frac{5}{\sqrt{1-(5x)^2}}\)
\(\,\,\,\,\,f'(x)=\displaystyle-\frac{5}{\sqrt{1-25x^2}}\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=\displaystyle-\frac{5}{\sqrt{1-25x^2}}\)
\(\textbf{13)}\) \(f(x)=\arctan(x^3), \text{find } f'(x)\)
The derivative is \(f'(x)=\displaystyle\frac{3x^2}{1+x^6}\)
\(\,\,\,\,\,f(x)=\arctan(x^3)\)
\(\,\,\,\,\,u=x^3,\quad u’=3x^2\)
\(\,\,\,\,\,\frac{d}{dx}\arctan(u)=\displaystyle\frac{u’}{1+u^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{3x^2}{1+(x^3)^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{3x^2}{1+x^6}\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=\displaystyle\frac{3x^2}{1+x^6}\)
\(\textbf{14)}\) \(f(x)=\arcsec(4x), \text{find } f'(x)\)
The derivative is \(f'(x)=\displaystyle\frac{1}{|x|\sqrt{16x^2-1}}\)
\(\,\,\,\,\,f(x)=\arcsec(4x)\)
\(\,\,\,\,\,u=4x,\quad u’=4\)
\(\,\,\,\,\,\frac{d}{dx}\arcsec(u)=\displaystyle\frac{u’}{|u|\sqrt{u^2-1}}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{4}{|4x|\sqrt{(4x)^2-1}}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{4}{4|x|\sqrt{16x^2-1}}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{1}{|x|\sqrt{16x^2-1}}\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=\displaystyle\frac{1}{|x|\sqrt{16x^2-1}}\)
\(\textbf{15)}\) \(f(x)=\arccsc(3x^2), \text{find } f'(x)\)
The derivative is \(f'(x)=\displaystyle-\frac{6x}{|3x^2|\sqrt{9x^4-1}}\)
\(\,\,\,\,\,f(x)=\arccsc(3x^2)\)
\(\,\,\,\,\,u=3x^2,\quad u’=6x\)
\(\,\,\,\,\,\frac{d}{dx}\arccsc(u)=\displaystyle-\frac{u’}{|u|\sqrt{u^2-1}}\)
\(\,\,\,\,\,f'(x)=\displaystyle-\frac{6x}{|3x^2|\sqrt{(3x^2)^2-1}}\)
\(\,\,\,\,\,f'(x)=\displaystyle-\frac{6x}{|3x^2|\sqrt{9x^4-1}}\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=\displaystyle-\frac{6x}{|3x^2|\sqrt{9x^4-1}}\)
\(\textbf{16)}\) \(f(x)=x^2\arctan(x), \text{find } f'(x)\)
The derivative is \(f'(x)=2x\arctan(x)+\displaystyle\frac{x^2}{1+x^2}\)
\(\,\,\,\,\,f(x)=x^2\arctan(x)\)
\(\,\,\,\,\,\text{Use the product rule.}\)
\(\,\,\,\,\,\frac{d}{dx}\left[x^2\right]=2x\)
\(\,\,\,\,\,\frac{d}{dx}\arctan(x)=\displaystyle\frac{1}{1+x^2}\)
\(\,\,\,\,\,f'(x)=2x\arctan(x)+x^2\left(\frac{1}{1+x^2}\right)\)
\(\,\,\,\,\,f'(x)=2x\arctan(x)+\displaystyle\frac{x^2}{1+x^2}\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=2x\arctan(x)+\displaystyle\frac{x^2}{1+x^2}\)
\(\textbf{17)}\) \(f(x)=\displaystyle\frac{\arctan(x)}{x^2+1}, \text{find } f'(x)\)
The derivative is \(f'(x)=\displaystyle\frac{1-2x\arctan(x)}{(x^2+1)^2}\)
\(\,\,\,\,\,f(x)=\displaystyle\frac{\arctan(x)}{x^2+1}\)
\(\,\,\,\,\,\text{Use the quotient rule.}\)
\(\,\,\,\,\,\frac{d}{dx}\arctan(x)=\displaystyle\frac{1}{1+x^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{(x^2+1)\left(\frac{1}{1+x^2}\right)-\arctan(x)(2x)}{(x^2+1)^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{1-2x\arctan(x)}{(x^2+1)^2}\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=\displaystyle\frac{1-2x\arctan(x)}{(x^2+1)^2}\)
\(\textbf{18)}\) \(f(x)=\arcsin(x^2-1), \text{find } f'(x)\)
The derivative is \(f'(x)=\displaystyle\frac{2x}{\sqrt{1-(x^2-1)^2}}\)
\(\,\,\,\,\,f(x)=\arcsin(x^2-1)\)
\(\,\,\,\,\,u=x^2-1,\quad u’=2x\)
\(\,\,\,\,\,\frac{d}{dx}\arcsin(u)=\displaystyle\frac{u’}{\sqrt{1-u^2}}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{2x}{\sqrt{1-(x^2-1)^2}}\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=\displaystyle\frac{2x}{\sqrt{1-(x^2-1)^2}}\)
\(\textbf{19)}\) \(f(x)=\arccos(e^x), \text{find } f'(x)\)
The derivative is \(f'(x)=\displaystyle-\frac{e^x}{\sqrt{1-e^{2x}}}\)
\(\,\,\,\,\,f(x)=\arccos(e^x)\)
\(\,\,\,\,\,u=e^x,\quad u’=e^x\)
\(\,\,\,\,\,\frac{d}{dx}\arccos(u)=\displaystyle-\frac{u’}{\sqrt{1-u^2}}\)
\(\,\,\,\,\,f'(x)=\displaystyle-\frac{e^x}{\sqrt{1-(e^x)^2}}\)
\(\,\,\,\,\,f'(x)=\displaystyle-\frac{e^x}{\sqrt{1-e^{2x}}}\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=\displaystyle-\frac{e^x}{\sqrt{1-e^{2x}}}\)
\(\textbf{20)}\) \(f(x)=\arctan(\sqrt{x}), \text{find } f'(x)\)
The derivative is \(f'(x)=\displaystyle\frac{1}{2\sqrt{x}(1+x)}\)
\(\,\,\,\,\,f(x)=\arctan(\sqrt{x})\)
\(\,\,\,\,\,u=\sqrt{x}=x^{1/2},\quad u’=\frac{1}{2\sqrt{x}}\)
\(\,\,\,\,\,\frac{d}{dx}\arctan(u)=\displaystyle\frac{u’}{1+u^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{\frac{1}{2\sqrt{x}}}{1+(\sqrt{x})^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{1}{2\sqrt{x}(1+x)}\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=\displaystyle\frac{1}{2\sqrt{x}(1+x)}\)
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