The chain rule is used to find derivatives of composite functions, where one function is inside another function. This page focuses on recognizing the outside and inside functions, differentiating the outside first, and then multiplying by the derivative of the inside. The practice problems include powers, radicals, trigonometric functions, exponential functions, logarithms, and table-based chain rule questions.
Notes
Practice Problems
Find the derivative of each
\(\small{\textbf{1)}}\) Find \(f'(x)\) where \(f(x)=(3x+5)^4\)
The derivative is \(f'(x)=12(3x+5)^3\)
\(\,\,\,\,\,\,f(x)=(3x+5)^4\)
\(\,\,\,\,\,\,f'(x)=\left(([3x+5])^4\right)’ \cdot (3x+5)’\)
\(\,\,\,\,\,\,f'(x)=4(3x+5)^3 \cdot 3\)
\(\,\,\,\,\,\,\)The derivative is \(f'(x)=12(3x+5)^3\)
\(\small{\textbf{2)}}\) Find \(f'(x)\) where \(f(x)=\sqrt{2x-1}\)
The derivative is \(f'(x)=\displaystyle\frac{1}{\sqrt{2x-1}}\)
\(\,\,\,\,\,\,f(x)=\sqrt{2x-1}\)
\(\,\,\,\,\,\,\displaystyle f(x)=\left(2x-1\right)^{1/2}\)
\(\,\,\,\,\,\,\displaystyle f'(x)=\left(\left([2x-1]\right)^{1/2}\right)’ \cdot (2x-1)’\)
\(\,\,\,\,\,\, f'(x)=\left(\frac{1}{2}\left(2x-1\right)^{-1/2}\right) \cdot (2)\)
\(\,\,\,\,\,\, f'(x)=\frac{2}{2}\left(2x-1\right)^{-1/2}\)
\(\,\,\,\,\,\, f'(x)=\frac{1}{1}\left(2x-1\right)^{-1/2}\)
\(\,\,\,\,\,\,\displaystyle f'(x)=\frac{1}{\sqrt{2x-1}}\)
\(\,\,\,\,\,\,\)The derivative is \(f'(x)=\displaystyle\frac{1}{\sqrt{2x-1}}\)
\(\small{\textbf{3)}}\) Find \(f'(x)\) where \(f(x)=\displaystyle\frac{1}{x^3+3x-2}\)
The derivative is \(f'(x)=-\displaystyle\frac{3x^2+3}{(x^3+3x-2)^2}\)
\(\,\,\,\,\,\,f(x)=\displaystyle\frac{1}{x^3+3x-2}\)
\(\,\,\,\,\,\,f(x)=\displaystyle \left(x^3+3x-2\right)^{-1}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \left(\left(\left[x^3+3x-2\right]\right)^{-1}\right)’ \cdot \left(x^3+3x-2\right)’\)
\(\,\,\,\,\,\,f'(x)=\displaystyle -1\left(\left[x^3+3x-2\right]\right)^{-2} \cdot \left(3x^2+3\right)\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \frac{-1}{\left(x^3+3x-2\right)^{2}} \cdot \left(3x^2+3\right)\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \frac{-1\left(3x^2+3\right)}{\left(x^3+3x-2\right)^{2}} \)
\(\,\,\,\,\,\,\)The derivative is \(f'(x)=-\displaystyle\frac{3x^2+3}{(x^3+3x-2)^2}\)
\(\small{\textbf{4)}}\) Find \(f'(x)\) where \(f(x)=x^2(2x-1)^3\)
\(f'(x)=2x(2x-1)^3+6x^2(2x-1)^2\) or \(f'(x)=2x(2x-1)^2(5x-1)\)
\(\,\,\,\,\,\,f(x)=x^2(2x-1)^3\)
\(\,\,\,\,\,\,f'(x)=\left(x^2\right)’ \cdot \left((2x-1)^3\right)+\left(x^2\right) \cdot \left((2x-1)^3\right)’\)
\(\,\,\,\,\,\,f'(x)=\left(2x\right) \cdot \left((2x-1)^3\right)+\left(x^2\right) \cdot \left(\left(([2x-1])^3\right)’ \cdot \left(2x-1\right)’\right)\)
\(\,\,\,\,\,\,f'(x)=\left(2x\right) \cdot \left((2x-1)^3\right)+\left(x^2\right) \cdot \left(3([2x-1])^2 \cdot \left(2\right)\right)\)
\(\,\,\,\,\,\,f'(x)=2x(2x-1)^3+3x^2(2x-1)^2 \cdot 2\)
\(\,\,\,\,\,\,f'(x)=2x(2x-1)^3+6x^2(2x-1)^2\)
\(\,\,\,\,\,\,\)The answer is \(f'(x)=2x(2x-1)^3+6x^2(2x-1)^2\) or \(f'(x)=2x(2x-1)^2(5x-1)\)
\(\small{\textbf{5)}}\) Find \(f'(x)\) where \(f(x)=\sqrt[3]{x^4-2x+1}\)
The derivative is \(f'(x)=\displaystyle\frac{4x^3-2}{3(x^4-2x+1)^{2/3}}\)
\(\,\,\,\,\,\,f(x)=\sqrt[3]{x^4-2x+1}\)
\(\,\,\,\,\,\,f(x)=\left(x^4-2x+1\right)^{1/3}\)
\(\,\,\,\,\,\,f'(x)=1/3\left(x^4-2x+1\right)^{-2/3} \cdot \left(4x^3-2\right)\)
\(\,\,\,\,\,\,f'(x)=\displaystyle\frac{4x^3-2}{3(x^4-2x+1)^{2/3}}\)
\(\small{\textbf{6)}}\) Find \(f'(x)\) where \(f(x)=(x^3-2)^4\)
The derivative is \(f'(x)=12x^2(x^3-2)^3\)
\(\small{\textbf{7)}}\) Find \(f'(x)\) where \(f(x) = (3x^2 + 5)^6\)
The derivative is \(f'(x) = 36x(3x^2+5)^5\)
\(f(x) = (3x^2 + 5)^6\)
\(f'(x) = 6(3x^2 + 5)^5 \cdot 6x\)
\(f'(x) = 36x(3x^2 + 5)^5\)
\(\small{\textbf{8)}}\) Find \(f'(x)\) where \(f(x) = \displaystyle\frac{1}{\left( x^2 + 2x – 5 \right)^ 3}\)
The derivative is \(f'(x) =- \displaystyle\frac{6(x+1)}{(x^2+2x-5)^4}\)
\(\,\,\,\,\,f(x) = \displaystyle\frac{1}{\left( x^2 + 2x – 5 \right)^ 3}\)
\(\,\,\,\,\,f(x) = \left( x^2 + 2x – 5 \right)^{ – 3}\)
\(\,\,\,\,\,f'(x) = -3\left( x^2 + 2x – 5 \right)^{ – 4} \cdot \left(2x+2\right)\)
\(\,\,\,\,\,f'(x) =- \displaystyle\frac{3(2x+2)}{(x^2+2x-5)^4}\)
\(\,\,\,\,\,f'(x) =- \displaystyle\frac{6(x+1)}{(x^2+2x-5)^4}\)
\(\small{\textbf{9)}}\) Find \(f'(x)\) where \(f(x) = \sqrt[5]{{1 – x}}\)
The derivative is \(f'(x) = -\displaystyle\frac{1}{5\sqrt[5]{(1-x)^4}}\)
\(\,\,\,\,\,f(x) = \sqrt[5]{1 – x}\)
\(\,\,\,\,\,f(x) = (1 – x)^{\frac{1}{5}}\)
\(\,\,\,\,\,f'(x) = \frac{1}{5}(1 – x)^{-\frac{4}{5}} \cdot (-1)\)
\(\,\,\,\,\,f'(x) = -\displaystyle\frac{1}{5\sqrt[5]{(1 – x)^4}}\)
\(\small{\textbf{10)}}\) Find \(f'(x)\) where \(f(x) = \csc \left( {2x} \right)\)
The hint is \( \frac{d}{dx} \csc x = -\csc x \cot x \)
The derivative is \(f'(x) = -2 \csc (2x) \cot(2x) \)
\(\small{\textbf{11)}}\) Find \(f'(x)\) where \(f(x) = 3\cos \left( {\tan \left( 4x \right)} \right)\)
The hint is \( \frac{d}{dx} \tan x = \sec^2 (x)\)
The derivative is \(f'(x) = -12 \sec^2(4x)\sin(\tan(4x))\)
\(\small{\textbf{12)}}\) Find \(f'(x)\) where \(f(x) = \tan \left( {15x + 14} \right)\)
The hint is \( \frac{d}{dx} \tan x = \sec^2 (x)\)
The derivative is \(f'(x) = 15 \sec^2(15x+14)\)
\(\small{\textbf{13)}}\) Find \(f'(x)\) where \(f(x) = e^{(x^2 + 2)}\)
The derivative is \(f'(x) = 2xe^{(x^2 + 2)}\)
\(\small{\textbf{14)}}\) Find \(f'(x)\) where \(f(x) = e^{1 – \sin \left( x \right)}\)
The derivative is \(f'(x) = -{e^{1 – \sin \left( x \right)}} \cos{(x)}\)
\(\small{\textbf{15)}}\) Find \(f'(x)\) where \(f(x) = {3^{1 + 4x}}\)
The derivative is \(f'(x) = 4 \ln(3) (3^{1 + 4x})\)
\(\small{\textbf{16)}}\) Find \(f'(x)\) where \(f(x) = {\tan ^{ – 1}}\left( {5x + 2} \right)\)
The hint is \( \frac{d}{dx} \arctan x = \displaystyle\frac{1}{1+x^2}\)
The derivative is \(f'(x) = \displaystyle\frac{1}{5x^2+4x+1}\)
\(\small{\textbf{17)}}\) Find \(f'(x)\) where \(f(x) = \ln \left({2x^3 +3x +6} \right)\)
The derivative is \(f'(x) = \displaystyle\frac{6x^2+3}{2x^3+3x+6}\)
\(\small{\textbf{18)}}\) Find \(f'(x)\) where \(f(x) = \ln \left( {\sin \left( x \right) + \cos \left( x \right)} \right)\)
The derivative is \(f'(x) = \displaystyle\frac{\cos{(x)}-\sin{(x)}}{\cos{(x)}+\sin{(x)}}\)
\(\small{\textbf{19)}}\) Find \(f'(x)\) where \(f(x) = \ln \left( {\tan \left( x \right) + \cot \left( x \right)} \right)\)
The hint is \( \frac{d}{dx} \tan x = \sec^2 (x),\,\frac{d}{dx} \cot x = -\csc^2 (x) \)
The derivative is \(f'(x) = \displaystyle\frac{\sec^2{(x)}-\csc^2{(x)}}{\tan{(x)}+\cot{(x)}}\)
\(\small{\textbf{20)}}\) Find \(f'(x)\) where \(f(x) = {\sin ^2}\left( x \right) + \sin \left( {{x^2}} \right)\)
The derivative is \(f'(x) = 2 \sin{(x)}\cos{(x)} + 2x \cos{(x^2)}\)
\(\small{\textbf{21)}}\) Find \(f'(x)\) where \(f(x) = {e^{ – x}}+e^x\)
The derivative is \(f'(x) = e^x-e^{-x}\)
\(\small{\textbf{22)}}\) Find \(f'(x)\) where \(f(x) = 2x^3 + \left(4x^5 – 6x \right)^7\)
The derivative is \(f'(x) = 6x^2 + 7 \left(4x^5 – 6x \right)^6 \left(20x^4-6 \right)\)
\(\small{\textbf{23)}}\) Find \(f'(x)\) where \(f(x) = \ln \left( {\sin \left( x \right)} \right) – \sin \left( \ln x \right) \)
The derivative is \(f'(x) = \cot{(x)}-\displaystyle\frac{cos{( \ln{x})}}{x}\)
\(\small{\textbf{24)}}\) Find \(f'(x)\) where \(f(x) = x^2\ln(x^2)\)
The derivative is \(f'(x) = 2x \ln(x^2)+2x\)
\(\small{\textbf{25)}}\) Find \(f'(x)\) where \(f(x) = \sin \left( {2x} \right)\cos \left( {2x} \right)\)
The derivative is \(f'(x) = 2\left[\cos^2{(2x)}-\sin^2{(2x)}\right]\)
\(\small{\textbf{26)}}\) Find \(f'(x)\) where \( f(x) = \sin \left( {\cos \left( {2x} \right)} \right)\)
The derivative is \(f'(x) = -2 \cos{(\cos{(2x)})} \sin{(2x)}\)
\(\small{\textbf{27)}}\) Find \(f'(x)\) where \(f(x) = \displaystyle\frac{\sin {2x} }{\cos{2x}}\)
The hint is \( \frac{\sin x}{\cos x} =\tan x\)
The derivative is \(f'(x) = 2 \sec^2{(2x)}\)
\(\small{\textbf{28)}}\) Find \(f'(x)\) where \(f(x) = \sin \left( e^x \right)\)
The derivative is \(f'(x) = e^x \cos{(e^x)}\)
\(\small{\textbf{29)}}\) Find \(f'(x)\) where \(f(x) = \sqrt {\tan \left( {2x} \right)}\)
The hint is \( \frac{d}{dx} \tan x = \sec^2 (x)\)
The derivative is \(f'(x) = \displaystyle\frac{\sec^2{(2x)}}{\sqrt{tan{(2x)}}}\)
\(\small{\textbf{30)}}\) Find \(f'(x)\) where \(f(x) = {\tan ^3}\left( {{x^3}} \right)\)
The hint is \( \frac{d}{dx} \tan x = \sec^2 (x)\)
The derivative is \(f'(x) = 9x^2 \tan^2{(x^3)} \sec^2{(x^3)}\)
\(\small{\textbf{31)}}\) Find \(f'(x)\) where \(f(x)=\ln(\ln x)\)
The answer is \(f'(x)=\displaystyle\frac{1}{x \ln x }\)
\(\small{\textbf{32)}}\) Find \(f'(x)\) where \(f(x)=\displaystyle 3^{\sin^4(x^3)}\)
The answer is \(f'(x)=\displaystyle 3^{\sin^4 \left( x^3 \right)} (\ln 3) \cdot 4 \left( \sin (x^3) \right)^3 \cdot \left( \cos (x^3) \right) \cdot 3x^2 \)
\(\small{\textbf{33)}}\) Find \(f'(x)\) where \(f(x)=\displaystyle \sec{\left(\sin(x)\right)}\)
The answer is \(f'(x)=\displaystyle \sec{\left(\sin{(x)}\right)} \tan{\left(\sin{(x)}\right)} \cdot \cos{(x)}\)
\(\small{\textbf{34)}}\) Find \(f'(x)\) where \(f(x)=\displaystyle \ln{\sqrt{x}}\)
The answer is \(f'(x)=\displaystyle \frac{1}{2x}\)
\(\small{\textbf{35)}}\) Find \(f'(3)\) where \(f(x)=g(h(x))\)
| \(x\) |
\(1\) |
\(2\) |
\(3\) |
\(4\) |
| \(g(x)\) |
\(9\) |
\(4\) |
\(6\) |
\(2\) |
| \(g'(x)\) |
\(8\) |
\(3\) |
\(5\) |
\(2\) |
| \(h(x)\) |
\(1\) |
\(3\) |
\(4\) |
\(2\) |
| \(h'(x)\) |
\(4\) |
\(-2\) |
\(7\) |
\(3\) |
\(f'(3)=14\)
\(\,\,\,\,\,f(x)=g(h(x))\)
\(\,\,\,\,\,f'(x)=g'(h(x))*h'(x)\)
\(\,\,\,\,\,f'(3)=g'(h(3))*h'(3)\)
\(\,\,\,\,\,h(3)=4\)
\(\,\,\,\,\,f'(3)=g'(4)*h'(3)\)
\(\,\,\,\,\,f'(3)=2*7\)
\(\,\,\,\,\,f'(3)=14\)
\(\small{\textbf{36)}}\) Find \(f'(1)\) where \(f(x)=g(h(x))\)
| \(x\) |
\(1\) |
\(2\) |
\(3\) |
\(4\) |
| \(g(x)\) |
\(9\) |
\(4\) |
\(6\) |
\(2\) |
| \(g'(x)\) |
\(8\) |
\(3\) |
\(5\) |
\(2\) |
| \(h(x)\) |
\(1\) |
\(3\) |
\(4\) |
\(2\) |
| \(h'(x)\) |
\(4\) |
\(-2\) |
\(7\) |
\(3\) |
\(f'(1)=32\)
\(\,\,\,\,\,f(x)=g(h(x))\)
\(\,\,\,\,\,f'(x)=g'(h(x))*h'(x)\)
\(\,\,\,\,\,f'(1)=g'(h(1))*h'(1)\)
\(\,\,\,\,\,h(1)=1\)
\(\,\,\,\,\,f'(1)=g'(1)*h'(1)\)
\(\,\,\,\,\,f'(1)=8*4\)
\(\,\,\,\,\,f'(1)=32\)
\(\small{\textbf{37)}}\) Find \(f'(2)\) where \(f(x)=g(h(x))\)
| \(x\) |
\(1\) |
\(2\) |
\(3\) |
\(4\) |
| \(g(x)\) |
\(9\) |
\(4\) |
\(6\) |
\(2\) |
| \(g'(x)\) |
\(8\) |
\(3\) |
\(5\) |
\(2\) |
| \(h(x)\) |
\(1\) |
\(3\) |
\(4\) |
\(2\) |
| \(h'(x)\) |
\(4\) |
\(-2\) |
\(7\) |
\(3\) |
\(f'(2)=-10\)
\(\,\,\,\,\,f(x)=g(h(x))\)
\(\,\,\,\,\,f'(x)=g'(h(x))*h'(x)\)
\(\,\,\,\,\,f'(2)=g'(h(2))*h'(2)\)
\(\,\,\,\,\,h(2)=3\)
\(\,\,\,\,\,f'(2)=g'(3)*h'(2)\)
\(\,\,\,\,\,f'(2)=5*(-2)\)
\(\,\,\,\,\,f'(2)=-10\)
\(\small{\textbf{38)}}\) Find \(f'(4)\) where \(f(x)=g(h(x))\)
| \(x\) |
\(1\) |
\(2\) |
\(3\) |
\(4\) |
| \(g(x)\) |
\(9\) |
\(4\) |
\(6\) |
\(2\) |
| \(g'(x)\) |
\(8\) |
\(3\) |
\(5\) |
\(2\) |
| \(h(x)\) |
\(1\) |
\(3\) |
\(4\) |
\(2\) |
| \(h'(x)\) |
\(4\) |
\(-2\) |
\(7\) |
\(3\) |
\(f'(4)=9\)
\(\,\,\,\,\,f(x)=g(h(x))\)
\(\,\,\,\,\,f'(x)=g'(h(x))*h'(x)\)
\(\,\,\,\,\,f'(4)=g'(h(4))*h'(4)\)
\(\,\,\,\,\,h(4)=2\)
\(\,\,\,\,\,f'(4)=g'(2)*h'(4)\)
\(\,\,\,\,\,f'(4)=3*3\)
\(\,\,\,\,\,f'(4)=9\)
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