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 \(f'(x)=-2\sin(x)\cos(x)+\displaystyle \frac{3}{\sqrt{1-x^2}}\)
\(\,\,\,\,\,f(x)=\cos^2(x)+3\sin^{-1}(x)\)
\(\,\,\,\,\,\frac{d}{dx}\left[\cos^2(x)\right]=2\cos(x)(-\sin(x))\)
\(\,\,\,\,\,\frac{d}{dx}\left[3\sin^{-1}(x)\right]=3\cdot\frac{1}{\sqrt{1-x^2}}\)
\(\,\,\,\,\,f'(x)=-2\sin(x)\cos(x)+\displaystyle \frac{3}{\sqrt{1-x^2}}\)
\(\,\,\,\,\,\)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 \(f'(x)=\displaystyle \frac{8}{\sqrt{1-x^2}}+\frac{2}{|x|\sqrt{x^2-1}}\)
\(\,\,\,\,\,f(x)=8\sin^{-1}(x)-2\csc^{-1}(x)\)
\(\,\,\,\,\,\frac{d}{dx}\sin^{-1}(x)=\displaystyle\frac{1}{\sqrt{1-x^2}}\)
\(\,\,\,\,\,\frac{d}{dx}\csc^{-1}(x)=\displaystyle-\frac{1}{|x|\sqrt{x^2-1}}\)
\(\,\,\,\,\,f'(x)=8\left(\frac{1}{\sqrt{1-x^2}}\right)-2\left(-\frac{1}{|x|\sqrt{x^2-1}}\right)\)
\(\,\,\,\,\,f'(x)=\displaystyle \frac{8}{\sqrt{1-x^2}}+\frac{2}{|x|\sqrt{x^2-1}}\)
\(\,\,\,\,\,\)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 \(f'(x)=\displaystyle\frac{2}{4x^2+1}+12\sec ^2\left(3x\right)\)
\)
\(\,\,\,\,\,f(x)=\arctan(2x)+4\tan(3x)\)
\(\,\,\,\,\,\frac{d}{dx}\arctan(u)=\displaystyle\frac{u’}{1+u^2}\)
\(\,\,\,\,\,u=2x,\quad u’=2\)
\(\,\,\,\,\,\frac{d}{dx}\arctan(2x)=\displaystyle\frac{2}{1+(2x)^2}\)
\(\,\,\,\,\,\frac{d}{dx}\left[4\tan(3x)\right]=4\cdot3\sec^2(3x)\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{2}{4x^2+1}+12\sec ^2\left(3x\right)\)
\(\,\,\,\,\,\)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 \(f'(x)=\displaystyle\frac{1}{|x|\sqrt{x^2-1}}+\frac{1}{\sqrt{1-x^2}}\)
\(\,\,\,\,\,f(x)=\sec^{-1}(x)-\cos^{-1}(x)\)
\(\,\,\,\,\,\frac{d}{dx}\sec^{-1}(x)=\displaystyle\frac{1}{|x|\sqrt{x^2-1}}\)
\(\,\,\,\,\,\frac{d}{dx}\cos^{-1}(x)=\displaystyle-\frac{1}{\sqrt{1-x^2}}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{1}{|x|\sqrt{x^2-1}}-\left(-\frac{1}{\sqrt{1-x^2}}\right)\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{1}{|x|\sqrt{x^2-1}}+\frac{1}{\sqrt{1-x^2}}\)
\(\,\,\,\,\,\)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 \(f'(x)=3x^2\arcsin \left(2x\right)+\displaystyle\frac{2x^3}{\sqrt{1-4x^2}}\)
\(\,\,\,\,\,f(x)=x^3\arcsin(2x)\)
\(\,\,\,\,\,\text{Use the product rule.}\)
\(\,\,\,\,\,\frac{d}{dx}\left[x^3\right]=3x^2\)
\(\,\,\,\,\,\frac{d}{dx}\arcsin(2x)=\displaystyle\frac{2}{\sqrt{1-(2x)^2}}\)
\(\,\,\,\,\,f'(x)=3x^2\arcsin(2x)+x^3\left(\frac{2}{\sqrt{1-4x^2}}\right)\)
\(\,\,\,\,\,f'(x)=3x^2\arcsin \left(2x\right)+\displaystyle\frac{2x^3}{\sqrt{1-4x^2}}\)
\(\,\,\,\,\,\)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 \(f'(x)=\displaystyle\frac{\frac{x}{\sqrt{1-x^2}}-\sin^{-1}(x)}{x^2}\)
\(\,\,\,\,\,f(x)=\displaystyle \frac{\sin^{-1}(x)}{x}\)
\(\,\,\,\,\,\text{Use the quotient rule.}\)
\(\,\,\,\,\,\frac{d}{dx}\sin^{-1}(x)=\displaystyle\frac{1}{\sqrt{1-x^2}}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{x\left(\frac{1}{\sqrt{1-x^2}}\right)-\sin^{-1}(x)(1)}{x^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{\frac{x}{\sqrt{1-x^2}}-\sin^{-1}(x)}{x^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{x-\sin^{-1}(x)\sqrt{1-x^2}}{x^2\sqrt{1-x^2}}\)
\(\,\,\,\,\,\)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 \(f'(x)=15x^4+\displaystyle\frac{15x^4}{9x^{10}+1}\)
\(\,\,\,\,\,f(x)=3x^5+\tan^{-1}(3x^5)\)
\(\,\,\,\,\,\frac{d}{dx}\left[3x^5\right]=15x^4\)
\(\,\,\,\,\,u=3x^5,\quad u’=15x^4\)
\(\,\,\,\,\,\frac{d}{dx}\tan^{-1}(u)=\displaystyle\frac{u’}{1+u^2}\)
\(\,\,\,\,\,\frac{d}{dx}\tan^{-1}(3x^5)=\displaystyle\frac{15x^4}{1+(3x^5)^2}\)
\(\,\,\,\,\,f'(x)=15x^4+\displaystyle\frac{15x^4}{9x^{10}+1}\)
\(\,\,\,\,\,f'(x)=\displaystyle \frac{135x^{14}+30x^4}{9x^{10}+1}\)
\(\,\,\,\,\,\)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 \(f'(x) =\displaystyle \frac{1}{x\sqrt{1-\left(\ln x \right)^2}}\)
\(\,\,\,\,\,f(x)=\sin^{-1}(\ln x)\)
\(\,\,\,\,\,u=\ln x,\quad u’=\frac{1}{x}\)
\(\,\,\,\,\,\frac{d}{dx} \arcsin{u}=\displaystyle \frac{u’}{\sqrt{1-u^2}}\)
\(\,\,\,\,\,\frac{d}{dx} \arcsin{u}=\displaystyle \frac{\frac{1}{x}}{\sqrt{1-(\ln{x})^2}}\)
\(\,\,\,\,\,f'(x) =\displaystyle \frac{1}{x\sqrt{1-\left(\ln x \right)^2}}\)
\(\,\,\,\,\,\)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 \(f'(x)=\displaystyle \frac{1}{\arcsin \left(x\right)\sqrt{1-x^2}}\)
\(\,\,\,\,\,f(x)=\ln(\arcsin(x))\)
\(\,\,\,\,\,u=\arcsin(x)\)
\(\,\,\,\,\,\frac{d}{dx}\ln(u)=\displaystyle\frac{u’}{u}\)
\(\,\,\,\,\,u’=\displaystyle\frac{1}{\sqrt{1-x^2}}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{\frac{1}{\sqrt{1-x^2}}}{\arcsin(x)}\)
\(\,\,\,\,\,f'(x)=\displaystyle \frac{1}{\arcsin \left(x\right)\sqrt{1-x^2}}\)
\(\,\,\,\,\,\)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 \(f'(x)=\displaystyle-\frac{2}{4x^2+1}\)
\(\,\,\,\,\,f(x)=\cot^{-1}(2x)\)
\(\,\,\,\,\,u=2x,\quad u’=2\)
\(\,\,\,\,\,\frac{d}{dx}\cot^{-1}(u)=\displaystyle-\frac{u’}{1+u^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle-\frac{2}{1+(2x)^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle-\frac{2}{4x^2+1}\)
\(\,\,\,\,\,\)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 \(\displaystyle\frac{dy}{dx}=\displaystyle\frac{2x+4}{1+\left(x^2+4x\right)^2}\)
\(\,\,\,\,\,y=\tan^{-1}(x^2+4x)\)
\(\,\,\,\,\,u=x^2+4x,\quad u’=2x+4\)
\(\,\,\,\,\,\frac{d}{dx}\tan^{-1}(u)=\displaystyle\frac{u’}{1+u^2}\)
\(\,\,\,\,\,\displaystyle\frac{dy}{dx}=\displaystyle\frac{2x+4}{1+\left(x^2+4x\right)^2}\)
\(\,\,\,\,\,\)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 \(f'(x)=\displaystyle-\frac{5}{\sqrt{1-25x^2}}\)
\(\,\,\,\,\,f(x)=\arccos(5x)\)
\(\,\,\,\,\,u=5x,\quad u’=5\)
\(\,\,\,\,\,\frac{d}{dx}\arccos(u)=\displaystyle-\frac{u’}{\sqrt{1-u^2}}\)
\(\,\,\,\,\,f'(x)=\displaystyle-\frac{5}{\sqrt{1-(5x)^2}}\)
\(\,\,\,\,\,f'(x)=\displaystyle-\frac{5}{\sqrt{1-25x^2}}\)
\(\,\,\,\,\,\)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 \(f'(x)=\displaystyle\frac{3x^2}{1+x^6}\)
\(\,\,\,\,\,f(x)=\arctan(x^3)\)
\(\,\,\,\,\,u=x^3,\quad u’=3x^2\)
\(\,\,\,\,\,\frac{d}{dx}\arctan(u)=\displaystyle\frac{u’}{1+u^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{3x^2}{1+(x^3)^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{3x^2}{1+x^6}\)
\(\,\,\,\,\,\)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 \(f'(x)=\displaystyle\frac{1}{|x|\sqrt{16x^2-1}}\)
\(\,\,\,\,\,f(x)=\arcsec(4x)\)
\(\,\,\,\,\,u=4x,\quad u’=4\)
\(\,\,\,\,\,\frac{d}{dx}\arcsec(u)=\displaystyle\frac{u’}{|u|\sqrt{u^2-1}}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{4}{|4x|\sqrt{(4x)^2-1}}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{4}{4|x|\sqrt{16x^2-1}}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{1}{|x|\sqrt{16x^2-1}}\)
\(\,\,\,\,\,\)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 \(f'(x)=\displaystyle-\frac{6x}{|3x^2|\sqrt{9x^4-1}}\)
\(\,\,\,\,\,f(x)=\arccsc(3x^2)\)
\(\,\,\,\,\,u=3x^2,\quad u’=6x\)
\(\,\,\,\,\,\frac{d}{dx}\arccsc(u)=\displaystyle-\frac{u’}{|u|\sqrt{u^2-1}}\)
\(\,\,\,\,\,f'(x)=\displaystyle-\frac{6x}{|3x^2|\sqrt{(3x^2)^2-1}}\)
\(\,\,\,\,\,f'(x)=\displaystyle-\frac{6x}{|3x^2|\sqrt{9x^4-1}}\)
\(\,\,\,\,\,\)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 \(f'(x)=2x\arctan(x)+\displaystyle\frac{x^2}{1+x^2}\)
\(\,\,\,\,\,f(x)=x^2\arctan(x)\)
\(\,\,\,\,\,\text{Use the product rule.}\)
\(\,\,\,\,\,\frac{d}{dx}\left[x^2\right]=2x\)
\(\,\,\,\,\,\frac{d}{dx}\arctan(x)=\displaystyle\frac{1}{1+x^2}\)
\(\,\,\,\,\,f'(x)=2x\arctan(x)+x^2\left(\frac{1}{1+x^2}\right)\)
\(\,\,\,\,\,f'(x)=2x\arctan(x)+\displaystyle\frac{x^2}{1+x^2}\)
\(\,\,\,\,\,\)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 \(f'(x)=\displaystyle\frac{1-2x\arctan(x)}{(x^2+1)^2}\)
\(\,\,\,\,\,f(x)=\displaystyle\frac{\arctan(x)}{x^2+1}\)
\(\,\,\,\,\,\text{Use the quotient rule.}\)
\(\,\,\,\,\,\frac{d}{dx}\arctan(x)=\displaystyle\frac{1}{1+x^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{(x^2+1)\left(\frac{1}{1+x^2}\right)-\arctan(x)(2x)}{(x^2+1)^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{1-2x\arctan(x)}{(x^2+1)^2}\)
\(\,\,\,\,\,\)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 \(f'(x)=\displaystyle\frac{2x}{\sqrt{1-(x^2-1)^2}}\)
\(\,\,\,\,\,f(x)=\arcsin(x^2-1)\)
\(\,\,\,\,\,u=x^2-1,\quad u’=2x\)
\(\,\,\,\,\,\frac{d}{dx}\arcsin(u)=\displaystyle\frac{u’}{\sqrt{1-u^2}}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{2x}{\sqrt{1-(x^2-1)^2}}\)
\(\,\,\,\,\,\)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 \(f'(x)=\displaystyle-\frac{e^x}{\sqrt{1-e^{2x}}}\)
\(\,\,\,\,\,f(x)=\arccos(e^x)\)
\(\,\,\,\,\,u=e^x,\quad u’=e^x\)
\(\,\,\,\,\,\frac{d}{dx}\arccos(u)=\displaystyle-\frac{u’}{\sqrt{1-u^2}}\)
\(\,\,\,\,\,f'(x)=\displaystyle-\frac{e^x}{\sqrt{1-(e^x)^2}}\)
\(\,\,\,\,\,f'(x)=\displaystyle-\frac{e^x}{\sqrt{1-e^{2x}}}\)
\(\,\,\,\,\,\)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 \(f'(x)=\displaystyle\frac{1}{2\sqrt{x}(1+x)}\)
\(\,\,\,\,\,f(x)=\arctan(\sqrt{x})\)
\(\,\,\,\,\,u=\sqrt{x}=x^{1/2},\quad u’=\frac{1}{2\sqrt{x}}\)
\(\,\,\,\,\,\frac{d}{dx}\arctan(u)=\displaystyle\frac{u’}{1+u^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{\frac{1}{2\sqrt{x}}}{1+(\sqrt{x})^2}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{1}{2\sqrt{x}(1+x)}\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=\displaystyle\frac{1}{2\sqrt{x}(1+x)}\)
See Related Pages\(\)
\(\bullet\text{ Inverse Trig Derivative Calculator }\)
\(\,\,\,\,\,\,\,\,\text{(Symbolab.com)}\)
\(\bullet\text{ Calculus Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)
\(\bullet\text{ Definition of Derivative}\)
\(\,\,\,\,\,\,\,\, \displaystyle \lim_{\Delta x\to 0} \frac{f(x+ \Delta x)-f(x)}{\Delta x} \)
\(\bullet\text{ Equation of the Tangent Line}\)
\(\,\,\,\,\,\,\,\,f(x)=x^3+3x^2−x \text{ at the point } (2,18)\)
\(\bullet\text{ Derivatives- Constant Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(c)=0\)
\(\bullet\text{ Derivatives- Power Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(x^n)=nx^{n-1}\)
\(\bullet\text{ Derivatives- Constant Multiple Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(cf(x))=cf'(x)\)
\(\bullet\text{ Derivatives- Sum and Difference Rules}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(x) \pm g(x)]=f'(x) \pm g'(x)\)
\(\bullet\text{ Derivatives- Sin and Cos}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}sin(x)=cos(x)\)
\(\bullet\text{ Derivatives- Product Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(x) \cdot g(x)]=f(x) \cdot g'(x)+f'(x) \cdot g(x)\)
\(\bullet\text{ Derivatives- Quotient Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\displaystyle\frac{f(x)}{g(x)}\right]=\displaystyle\frac{g(x) \cdot f'(x)-f(x) \cdot g'(x)}{[g(x)]^2}\)
\(\bullet\text{ Derivatives- Chain Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(g(x))]= f'(g(x)) \cdot g'(x)\)
\(\bullet\text{ Derivatives- ln(x)}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[ln(x)]= \displaystyle \frac{1}{x}\)
\(\bullet\text{ Implicit Differentiation}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Horizontal Tangent Line}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Mean Value Theorem}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Related Rates}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Increasing and Decreasing Intervals}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Intervals of concave up and down}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Inflection Points}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Graph of f(x), f'(x) and f”(x)}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Newton’s Method}\)
\(\,\,\,\,\,\,\,\,x_{n+1}=x_n – \displaystyle \frac{f(x_n)}{f'(x_n)}\)