The Chain Rule
Solved Problems
Example 11.
\[y = \ln \tan x\]
Solution.
This function exists for \(\pi n \lt x \lt \frac{\pi }{2} + \pi n,\) \(n \in \mathbb{Z}\). Its derivative is
\[y'\left( x \right) = \left( {\ln \tan x} \right)^\prime = \frac{1}{{\tan x}} \cdot {\left( {\tan x} \right)^\prime } = \frac{1}{{\tan x}} \cdot \frac{1}{{{{\cos }^2}x}} = \frac{1}{{\frac{{\sin x}}{{\cos x}}}} \cdot \frac{1}{{{{\cos }^2}x}} = \frac{{\cos x}}{{\sin x\,{{\cos }^2}x}} = \frac{1}{{\sin x\cos x}} = \frac{2}{{2\sin x\cos x}} = \frac{2}{{\sin 2x}}.\]
Example 12.
\[y = \sqrt {{x^2} + 3}\]
Solution.
The outer function is \(y = f\left( u \right) = \sqrt u .\) The inner function is \(u = g\left( x \right) = {x^2} + 3.\)
Then the derivative of the composite function \(y = f\left( {g\left( x \right)} \right) = \sqrt {{x^2} + 3} \) is written as
\[y^\prime = \left( {\sqrt {{x^2} + 3} } \right)^\prime = \frac{1}{{2\sqrt {{x^2} + 3} }} \cdot \left( {{x^2} + 3} \right)^\prime = \frac{1}{{2\sqrt {{x^2} + 3} }} \cdot 2x = \frac{x}{{\sqrt {{x^2} + 3} }}.\]
Example 13.
\[y = {\left( {1 + 3x} \right)^2}\]
Solution.
The outer function is the quadratic function \(y = {u^2},\) the inner function is the linear function \(u = 1 + 3x.\) By the chain rule,
\[y^\prime = \left[ {{{\left( {1 + 3x} \right)}^2}} \right]^\prime = 2\left( {1 + 3x} \right) \cdot \left( {1 + 3x} \right)^\prime = \left( {2 + 6x} \right) \cdot 3 = 6 + 18x.\]
Example 14.
\[y = \sqrt {{x^2} + 2x + 3} \]
Solution.
Using the formula for the derivative of the square root and the chain rule rule, we obtain:
\[y'\left( x \right) = \left( {\sqrt {{x^2} + 2x + 3} } \right)^\prime = \frac{1}{{2\sqrt {{x^2} + 2x + 3} }} \cdot {\left( {{x^2} + 2x + 3} \right)^\prime } = \frac{{2x + 2}}{{2\sqrt {{x^2} + 2x + 3} }} = \frac{{2\left( {x + 1} \right)}}{{2\sqrt {{x^2} + 2x + 3} }} = \frac{{x + 1}}{{\sqrt {{x^2} + 2x + 3} }}.\]
Example 15.
\[y = {\tan ^2}\frac{x}{2}\]
Solution.
Applying the chain rule twice we have
\[y'\left( x \right) = \left( {{{\tan }^2}\frac{x}{2}} \right)^\prime = 2\tan \frac{x}{2} \cdot {\left( {\tan \frac{x}{2}} \right)^\prime } = 2\tan \frac{x}{2} \cdot \frac{1}{{{{\cos }^2}\frac{x}{2}}} \cdot \frac{1}{2} = \tan \frac{x}{2}{\sec ^2}\frac{x}{2}.\]
The original function and its derivative are defined for
\[\frac{x}{2} \ne \frac{\pi }{2} + \pi n\;\;\;\text{or}\;\;x \ne \pi + 2\pi n, \;n \in \mathbb{Z}.\]
Example 16.
\[y = \ln \ln x\]
Solution.
\[y'\left( x \right) = \left( {\ln \ln x} \right)^\prime = \frac{1}{{\ln x}} \cdot {\left( {\ln x} \right)^\prime } = \frac{1}{{\ln x}} \cdot \frac{1}{x} = \frac{1}{{x\ln x}}.\]
In this example, the domain of the function and its derivative is determined by the condition \(\ln x \gt 0\) or \(x \gt 1.\)
Example 17.
\[y = {\left( {x\ln x} \right)^2}\]
Solution.
This example combines the chain rule and the product rule:
\[y^\prime = \left[ {{{\left( {x\ln x} \right)}^2}} \right]^\prime = 2\left( {x\ln x} \right) \cdot \left( {x\ln x} \right)^\prime = 2x\ln x \cdot \left[ {x^\prime\ln x + x\left( {\ln x} \right)^\prime} \right] = 2x\ln x \cdot \left[ {\ln x + x \cdot \frac{1}{x}} \right] = 2x\ln x\left( {\ln x + 1} \right).\]
Example 18.
\[y = {\left( {\sqrt x - 2} \right)^7}\]
Solution.
\[y'\left( x \right) = \left[ {{{\left( {\sqrt x - 2} \right)}^7}} \right]^\prime = 7{\left( {\sqrt x - 2} \right)^6} \cdot {\left( {\sqrt x - 2} \right)^\prime } = 7{\left( {\sqrt x - 2} \right)^6} \cdot \frac{1}{{2\sqrt x }} = \frac{{7{{\left( {\sqrt x - 2} \right)}^6}}}{{2\sqrt x }}\;\;\left( {x \gt 0} \right).\]
Example 19.
\[y = {\log _3}{x^2}\]
Solution.
Using the standard formula for the derivative of the logarithmic function
\[\left( {{{\log }_a}x} \right)^\prime = \frac{1}{{x\ln a}}\]
and applying the chain rule, we obtain:
\[y'\left( x \right) = \left( {{{\log }_3}{x^2}} \right)^\prime = \frac{1}{{{x^2}\ln 3}} \cdot {\left( {{x^2}} \right)^\prime } = \frac{{2x}}{{{x^2}\ln 3}} = \frac{2}{{x\ln 3}}\;\left( {x \ne 0} \right).\]
Example 20.
\[y = \ln \frac{1}{{{x^2}}}\]
Solution.
The outer function is the natural logarithm \(y = \ln u,\) the inner function is \(u = \frac{1}{{{x^2}}} = {x^{ - 2}}.\) Then
\[y^\prime = \left( {\ln \frac{1}{{{x^2}}}} \right)^\prime = \frac{1}{{\frac{1}{{{x^2}}}}} \cdot \left( {\frac{1}{{{x^2}}}} \right)^\prime = {x^2} \cdot \left( {{x^{ - 2}}} \right)^\prime = {x^2} \cdot \left( { - 2{x^{ - 3}}} \right) = - \frac{{2{x^2}}}{{{x^3}}} = - \frac{2}{x}.\]
Example 21.
\[y = {e^{\sin x}}\]
Solution.
Using the formula for the derivative of the exponential function
\[\left( {{e^x}} \right)^\prime = {e^x}\]
and differentiating as a composite function, we get
\[{\left( {{e^{\sin x}}} \right)^\prime = e^{\sin x}} \cdot {\left( {\sin x} \right)^\prime = e^{\sin x}}\cos x.\]
Example 22.
\[y = \sqrt {3{x^2} + 1} \]
Solution.
\[y'\left( x \right) = \left( {\sqrt {3{x^2} + 1} } \right)^\prime = \frac{1}{{2\sqrt {3{x^2} + 1} }} \cdot {\left( {3{x^2} + 1} \right)^\prime } = \frac{{6x}}{{2\sqrt {3{x^2} + 1} }} = \frac{{3x}}{{\sqrt {3{x^2} + 1} }}.\]
Example 23.
Find the value of the derivative of the function \(y = \sin {\frac{x}{3}}\) at \(x = 2\pi.\)
Solution.
The derivative of the function is given by
\[y'\left( x \right) = \left( {\sin \frac{x}{3}} \right)^\prime = \cos \frac{x}{3} \cdot {\left( {\frac{x}{3}} \right)^\prime } = \frac{1}{3}\cos \frac{x}{3}.\]
Substituting the value \(x = 2\pi,\) we obtain:
\[y'\left( {2\pi } \right) = \frac{1}{3}\cos \frac{{2\pi }}{3} = \frac{1}{3} \cdot \left( { - \frac{1}{2}} \right) = - \frac{1}{6}.\]
Example 24.
\[y = \frac{1}{a}\arctan \frac{x}{a}\]
Solution.
It is known that the derivative of the inverse tangent function is
\[\left( {\arctan x} \right)^\prime = \frac{1}{{1 + {x^2}}}.\]
Hence,
\[y'\left( x \right) = \left( {\frac{1}{a}\arctan \frac{x}{a}} \right)^\prime = \frac{1}{a}{\left( {\arctan \frac{x}{a}} \right)^\prime } = \frac{1}{a} \cdot \frac{1}{{1 + {{\left( {\frac{x}{a}} \right)}^2}}} \cdot {\left( {\frac{x}{a}} \right)^\prime } = \frac{1}{a} \cdot \frac{1}{{1 + {{\left( {\frac{x}{a}} \right)}^2}}} \cdot \frac{1}{a} = \frac{1}{{{a^2}}} \cdot \frac{1}{{\frac{{{a^2} + {x^2}}}{{{a^2}}}}} = \frac{{\cancel{a^2}}}{{\cancel{a^2}\left( {{x^2} + {a^2}} \right)}} = \frac{1}{{{x^2} + {a^2}}}.\]
Example 25.
\[y = \tan \frac{x}{2} - \cot \frac{x}{2}\]
Solution.
\[y'\left( x \right) = \left( {\tan \frac{x}{2} - \cot \frac{x}{2}} \right)^\prime = \left( {\tan \frac{x}{2}} \right)^\prime - \left( {\cot \frac{x}{2}} \right)^\prime = \frac{1}{{{{\cos }^2}\frac{x}{2}}} \cdot {\left( {\frac{x}{2}} \right)^\prime } = \left( { - \frac{1}{{{{\sin }^2}\frac{x}{2}}}} \right) \cdot {\left( {\frac{x}{2}} \right)^\prime } = \frac{1}{{2\,{{\cos }^2}\frac{x}{2}}} + \frac{1}{{2\,{{\sin }^2}\frac{x}{2}}} = \frac{{{{\sin }^2}\frac{x}{2} + {{\cos }^2}\frac{x}{2}}}{{2\,{{\sin }^2}\frac{x}{2}{\cos^2}\frac{x}{2}}}.\]
Next, we can use the Pythagorean trigonometric identity in the numerator and simplify the denominator by the double angle formula:
\[y'\left( x \right) = \frac{{{{\sin }^2}\frac{x}{2} + {{\cos }^2}\frac{x}{2}}}{{2\,{{\sin }^2}\frac{x}{2}{\cos^2}\frac{x}{2}}} = \frac{1}{{2\,{{\sin }^2}\frac{x}{2}{\cos^2}\frac{x}{2}}} = \frac{2}{{{{\left( {2\sin \frac{x}{2}\cos\frac{x}{2}} \right)}^2}}} = \frac{2}{{\sin x}}.\]
The domain of this function contains any values of \(x,\) except the following points: \(x \ne \pi n\), \(n \in \mathbb{Z}.\)
Example 26.
\[y = \sqrt {\sin 2x + 1} \]
Solution.
Using the chain rule twice, we have
\[y'\left( x \right) = \left( {\sqrt {\sin 2x + 1} } \right)^\prime = \frac{1}{{2\sqrt {\sin 2x + 1} }} \cdot {\left( {\sin 2x + 1} \right)^\prime } = \frac{1}{{2\sqrt {\sin 2x + 1} }} \cdot \cos 2x \cdot {\left( {2x} \right)^\prime } = \frac{{\cancel{2}\cos 2x}}{{\cancel{2}\sqrt {\sin 2x + 1} }} = \frac{{\cos 2x}}{{\sqrt {\sin 2x + 1} }}.\]
The derivative of this function is not defined at the points where the denominator is zero, that is when
\[\sin 2x + 1 = 0,\;\;\Rightarrow \sin 2x = - 1,\;\; \Rightarrow 2x = \frac{{3\pi }}{2} + 2\pi n,\;\;\Rightarrow x = \frac{{3\pi }}{4} + \pi n,\;n \in \mathbb{Z}.\]
Example 27.
\[y = \sin \left( {\ln \cos x} \right)\]
Solution.
This function exists for
\[\cos x \gt 0;\;\;\text{or}\;\; - \frac{\pi }{2} + 2\pi n \lt x \lt \frac{\pi }{2} + 2\pi n,\;n \in \mathbb{Z}.\]
Double application of the chain rule leads to the following expression for the derivative:
\[y'\left( x \right) = {\left[ {\sin \left( {\ln \cos x} \right)} \right]^\prime } = \cos \left( {\ln \cos x} \right) \cdot {\left( {\ln \cos x} \right)^\prime } = \cos \left( {\ln \cos x} \right) \cdot \frac{1}{{\cos x}} \cdot {\left( {\cos x} \right)^\prime } = \cos \left( {\ln \cos x} \right) \cdot \frac{1}{{\cos x}} \cdot \left( { - \sin x} \right) = \cos \left( {\ln \cos x} \right) \cdot \left( { - \tan x} \right) = - \tan x\cos \left( {\ln \cos x} \right).\]
Example 28.
\[y = x\sqrt {1 + {x^2}} \]
Solution.
First we differentiate as the product of functions:
\[y'\left( x \right) = \left( {x\sqrt {1 + {x^2}} } \right)^\prime = x' \cdot \sqrt {1 + {x^2}} + x \cdot {\left( {\sqrt {1 + {x^2}} } \right)^\prime }.\]
The latter derivative can be found by the chain rule:
\[y'\left( x \right) = \left( {x\sqrt {1 + {x^2}} } \right)^\prime = x' \cdot \sqrt {1 + {x^2}} + x \cdot {\left( {\sqrt {1 + {x^2}} } \right)^\prime } = 1 \cdot \sqrt {1 + {x^2}} + x \cdot \frac{1}{{2\sqrt {1 + {x^2}} }} \cdot {\left( {1 + {x^2}} \right)^\prime } = \sqrt {1 + {x^2}} + \frac{x}{{2\sqrt {1 + {x^2}} }} \cdot 2x = \sqrt {1 + {x^2}} + \frac{{\cancel{2}{x^2}}}{{\cancel{2}\sqrt {1 + {x^2}} }} = \frac{{{{\left( {\sqrt {1 + {x^2}} } \right)}^2} + {x^2}}}{{\sqrt {1 + {x^2}} }} = \frac{{1 + {x^2} + {x^2}}}{{\sqrt {1 + {x^2}} }} = \frac{{1 + 2{x^2}}}{{\sqrt {1 + {x^2}} }}.\]
Example 29.
\[y = \frac{1}{{\sqrt {{a^2} - {x^2}} }}\]
Solution.
We can rewrite the function as follows:
\[y = \frac{1}{{\sqrt {{a^2} - {x^2}} }} = \frac{1}{{{{\left( {{a^2} - {x^2}} \right)}^{\frac{1}{2}}}}} = {\left( {{a^2} - {x^2}} \right)^{ - \frac{1}{2}}}.\]
Using the chain rule, we have
\[y^\prime = \left[ {{{\left( {{a^2} - {x^2}} \right)}^{ - \frac{1}{2}}}} \right]^\prime = - \frac{1}{2}{\left( {{a^2} - {x^2}} \right)^{ - \frac{3}{2}}} \cdot \left( {{a^2} - {x^2}} \right)^\prime = - \frac{1}{{2{{\left( {{a^2} - {x^2}} \right)}^{\frac{3}{2}}}}} \cdot \left( { - 2x} \right) = \frac{{\cancel{2}x}}{{\cancel{2}\sqrt {{{\left( {{a^2} - {x^2}} \right)}^3}} }} = \frac{x}{{\sqrt {{{\left( {{a^2} - {x^2}} \right)}^3}} }}.\]
Example 30.
\[y = {\left( {\frac{{x + 1}}{{x - 1}}} \right)^3}\]
Solution.
Using the chain rule and the quotient rule, we have
\[y'\left( x \right) = \left[ {{{\left( {\frac{{x + 1}}{{x - 1}}} \right)}^3}} \right]^\prime = 3{\left( {\frac{{x + 1}}{{x - 1}}} \right)^2} \cdot {\left( {\frac{{x + 1}}{{x - 1}}} \right)^\prime } = 3{\left( {\frac{{x + 1}}{{x - 1}}} \right)^2} \cdot \frac{{1 \cdot \left( {x - 1} \right) - \left( {x + 1} \right) \cdot 1}}{{{{\left( {x - 1} \right)}^2}}} = 3{\left( {\frac{{x + 1}}{{x - 1}}} \right)^2} \cdot \frac{{\cancel{\color{blue}x} - \color{red}{1} - \cancel{\color{blue}x} - \color{red}{1}}}{{{{\left( {x - 1} \right)}^2}}} = 3{\left( {\frac{{x + 1}}{{x - 1}}} \right)^2} \cdot \frac{{\left( { - \color{red}{2}} \right)}}{{{{\left( {x - 1} \right)}^2}}} = - 6\frac{{{{\left( {x + 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^4}}}\;\;\left( {x \ne 1} \right).\]
Example 31.
\[y = {\left( {\frac{x}{{1 + x}}} \right)^4}\]
Solution.
Combining the chain, power and quotient rules, we have
\[y^\prime = \left[ {{{\left( {\frac{x}{{1 + x}}} \right)}^4}} \right]^\prime = 4{\left( {\frac{x}{{1 + x}}} \right)^3} \cdot \left( {\frac{x}{{1 + x}}} \right)^\prime = \frac{{4{x^3}}}{{{{\left( {1 + x} \right)}^3}}} \cdot \frac{{x^\prime\left( {1 + x} \right) - x\left( {1 + x} \right)^\prime}}{{{{\left( {1 + x} \right)}^2}}} = \frac{{4{x^3}}}{{{{\left( {1 + x} \right)}^3}}} \cdot \frac{{1 \cdot \left( {1 + x} \right) - x \cdot 1}}{{{{\left( {1 + x} \right)}^2}}} = \frac{{4{x^3}}}{{{{\left( {1 + x} \right)}^3}}} \cdot \frac{{1 + \cancel{x} - \cancel{x}}}{{{{\left( {1 + x} \right)}^2}}} = \frac{{4{x^3}}}{{{{\left( {1 + x} \right)}^5}}}.\]
Example 32.
\[y = \ln \left( {\sqrt {x + 1} - \sqrt x } \right)\]
Solution.
\[y'\left( x \right) = \left[ {\ln \left( {\sqrt {x + 1} - \sqrt x } \right)} \right]^\prime = \frac{1}{{\sqrt {x + 1} - \sqrt x }} \cdot {\left( {\sqrt {x + 1} - \sqrt x } \right)^\prime } = \frac{{\frac{1}{{2\sqrt {x + 1} }} - \frac{1}{{2\sqrt x }}}}{{\sqrt {x + 1} - \sqrt x }} = \frac{1}{2}\frac{{\frac{{\sqrt x - \sqrt {x + 1} }}{{\sqrt x \sqrt {x + 1} }}}}{{\sqrt {x + 1} - \sqrt x }} = - \frac{1}{2}\frac{\cancel{\sqrt {x + 1} - \sqrt x }}{{\sqrt x \sqrt {x + 1} \cancel{\left( {\sqrt {x + 1} - \sqrt x } \right)}}} = - \frac{1}{2}\frac{1}{{\sqrt {{x^2} + x} }}.\]
Example 33.
\[y = \sqrt[3]{{9{x^2} - 1}}\]
Solution.
Take into account that
\[{\left( {\sqrt[3]{x}} \right)^\prime } = \frac{1}{{3\sqrt[3]{{{x^2}}}}}.\]
Then
\[y'\left( x \right) = \left( {\sqrt[3]{{9{x^2} - 1}}} \right)^\prime = \frac{1}{{3\sqrt[3]{{{{\left( {9{x^2} - 1} \right)}^2}}}}} \cdot {\left( {9{x^2} - 1} \right)^\prime } = \frac{{18x}}{{3\sqrt[3]{{{{\left( {9{x^2} - 1} \right)}^2}}}}} = \frac{{6x}}{{\sqrt[3]{{{{\left( {9{x^2} - 1} \right)}^2}}}}}\;\left( {x \ne \pm \frac{1}{3}} \right).\]
Example 34.
\[y = {\left( {5x + 2} \right)^{13}} - {\left( {6x + 7} \right)^{10}}\]
Solution.
Applying the formula for the derivative of the difference of functions, the power rule and the chain rule, we obtain the following expression for the derivative:
\[y'\left( x \right) = \left[ {{{\left( {5x + 2} \right)}^{13} - {\left( {6x + 7} \right)}^{10}}} \right]^\prime = \left[ {{{\left( {5x + 2} \right)}^{13}}} \right]^\prime - \left[ {{{\left( {6x + 7} \right)}^{10}}} \right]^\prime = 13{\left( {5x + 2} \right)^{12} \cdot \left( {5x + 2} \right)^\prime } = 10{\left( {6x + 7} \right)^9 \cdot \left( {6x + 7} \right)^\prime } = 13{\left( {5x + 2} \right)^{12}} \cdot 5 - 10{\left( {6x + 7} \right)^9} \cdot 6 = 65{\left( {5x + 2} \right)^{12}} - 60{\left( {6x + 7} \right)^9}.\]
Example 35.
\[y = {\left( {x + \sqrt x } \right)^3}\]
Solution.
\[y'\left( x \right) = \left[ {{{\left( {x + \sqrt x } \right)}^3}} \right]^\prime = 3{\left( {x + \sqrt x } \right)^2} \cdot {\left( {x + \sqrt x } \right)^\prime } = 3{\left( {x + \sqrt x } \right)^2} \cdot \left( {1 + \frac{1}{{2\sqrt x }}} \right) = 3{\left( {x + \sqrt x } \right)^2} \cdot \frac{{2\sqrt x + 1}}{{2\sqrt x }} = \frac{{3{{\left( {\sqrt x } \right)}^2}{{\left( {\sqrt x + 1} \right)}^2}\left( {2\sqrt x + 1} \right)}}{{2\sqrt x }} = \frac{{3\sqrt x {{\left( {\sqrt x + 1} \right)}^2}\left( {2\sqrt x + 1} \right)}}{2}\;\;\left( {x \ge 0} \right).\]
Example 36.
\[y = \ln \left( {\frac{{x + 1}}{{x - 1}}} \right)\]
Solution.
By the chain rule and the quotient rule, we have
\[y'\left( x \right) = \left[ {\ln \left( {\frac{{x + 1}}{{x - 1}}} \right)} \right]^\prime = \frac{1}{{\frac{{x + 1}}{{x - 1}}}} \cdot {\left( {\frac{{x + 1}}{{x - 1}}} \right)^\prime } = \frac{{x - 1}}{{x + 1}} \cdot \frac{{1 \cdot \left( {x - 1} \right) - \left( {x + 1} \right) \cdot 1}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{x - 1}}{{x + 1}} \cdot \frac{{\cancel{\color{blue}x} - \color{red}{1} - \cancel{\color{blue}x} - \color{red}{1}}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{ - \color{red}{2}}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}.\]
