The Chain Rule
Solved Problems
Example 37.
\[y = \ln \frac{{3 - {x^2}}}{{2 - {x^2}}}\]
Solution.
Using the chain rule, the quotient rule and the power rule, we get:
\[\require{cancel} y^\prime = \left( {\ln \frac{{3 - {x^2}}}{{2 - {x^2}}}} \right)^\prime = \frac{1}{{\frac{{3 - {x^2}}}{{2 - {x^2}}}}} \cdot \left( {\frac{{3 - {x^2}}}{{2 - {x^2}}}} \right)^\prime = \frac{{2 - {x^2}}}{{3 - {x^2}}} \cdot \frac{{\left( { - 2x} \right) \left( {2 - {x^2}} \right) - \left( {3 - {x^2}} \right)\left( { - 2x} \right)}}{{{{\left( {2 - {x^2}} \right)}^2}}} = \frac{{ - 4x + \cancel{2{x^3}} + 6x - \cancel{2{x^3}}}}{{\left( {3 - {x^2}} \right)\left( {2 - {x^2}} \right)}} = \frac{{2x}}{{\left( {3 - {x^2}} \right)\left( {2 - {x^2}} \right)}} = \frac{{2x}}{{6 - 2{x^2} - 3{x^2} + {x^4}}} = \frac{{2x}}{{{x^4} - 5{x^2} + 6}}.\]
Example 38.
\[y = \sin {x^3}\cos {x^2}\]
Solution.
First differentiating by the product rule and then using the chain rule, we have
\[y'\left( x \right) = \left( {\sin {x^3}\cos {x^2}} \right)^\prime = {\left( {\sin {x^3}} \right)^\prime }\cos {x^2} + \sin {x^3}{\left( {\cos {x^2}} \right)^\prime } = \cos {x^3} \cdot {\left( {{x^3}} \right)^\prime } \cdot \cos {x^2} + \sin {x^3} \cdot \left( { - \sin{x^2}} \right) \cdot {\left( {{x^2}} \right)^\prime } = \cos {x^3} \cdot 3{x^2} \cdot \cos {x^2} - \sin{x^3} \cdot \sin{x^2} \cdot 2x = 3{x^2}\cos {x^3}\cos {x^2} - 2x \sin{x^3}\sin{x^2}.\]
Example 39.
\[y = \sin \left( {{{\cos }^2}x} \right)\]
Solution.
Differentiate twice as a composite function:
\[y'\left( x \right) = \left[ {\sin \left( {{{\cos }^2}x} \right)} \right]^\prime = \cos\left( {{{\cos }^2}x} \right) \cdot {\left( {{{\cos }^2}x} \right)^\prime } = \cos\left( {{{\cos }^2}x} \right) \cdot 2\cos x \cdot {\left( {\cos x} \right)^\prime } = \cos\left( {{{\cos }^2}x} \right) \cdot 2\cos x \cdot \left( { - \sin x} \right) = - \cos\left( {{{\cos }^2}x} \right) \cdot 2\cos x\sin x = - \cos\left( {{{\cos }^2}x} \right)\sin 2x.\]
Example 40.
\[y = \arcsin \frac{1}{x}\]
Solution.
The derivative of the inverse sine function is a standard derivative:
\[\left( {\arcsin x} \right)^\prime = \frac{1}{{\sqrt {1 - {x^2}} }}.\]
Using the chain rule we can write the following expression:
\[y'\left( x \right) = {\left( {\arcsin \frac{1}{x}} \right)^\prime } = \frac{1}{{\sqrt {1 - {{\left( {\frac{1}{x}} \right)}^2}} }} \cdot {\left( {\frac{1}{x}} \right)^\prime } = \frac{1}{{\sqrt {1 - \frac{1}{{{x^2}}}} }} \cdot \left( { - \frac{1}{{{x^2}}}} \right) = - \frac{1}{{{x^2}\sqrt {1 - \frac{1}{{{x^2}}}} }} = - \frac{1}{{{x^2}\sqrt {\frac{{{x^2} - 1}}{{{x^2}}}} }} = - \frac{{\sqrt {{x^2}} }}{{{x^2}\sqrt {{x^2} - 1} }}.\]
Take into account that \(\sqrt {{x^2}} = \left| x \right|\) and, hence, \({\left| x \right|^2} = {x^2}\). Then
\[y'\left( x \right) = - \frac{{\sqrt {{x^2}} }}{{{x^2}\sqrt {{x^2} - 1} }} = - \frac{{\left| x \right|}}{{{{\left| x \right|}^2}\sqrt {{x^2} - 1} }} = - \frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}.\]
The domain of the function and the derivative is given by \(x \in \left( { - \infty , - 1} \right) \cup \left( {1, + \infty } \right).\)
Example 41.
\[y = \sqrt {x\sqrt x } \]
Solution.
Using the product rule and the chain rule, we get:
\[y'\left( x \right) = {\left( {\sqrt {x\sqrt x } } \right)^\prime } = \frac{1}{{2\sqrt {x\sqrt x } }} \cdot {\left( {x\sqrt x } \right)^\prime } = \frac{1}{{2\sqrt {x\sqrt x } }} \cdot \left( {x'\sqrt x + x{{\left( {\sqrt x } \right)}^\prime }} \right) = \frac{1}{{2\sqrt {x\sqrt x } }} \cdot \left( {1 \cdot \sqrt x + x \cdot \frac{1}{{2\sqrt x }}} \right) = \frac{{\sqrt x + \frac{x}{{2\sqrt x }}}}{{2\sqrt {x\sqrt x } }} = \frac{{\sqrt x + \frac{{\sqrt x }}{2}}}{{2\sqrt {x\sqrt x } }} = \frac{{\frac{{3\sqrt x }}{2}}}{{2\sqrt {x\sqrt x } }} = \frac{{3 \cancel{\sqrt x} }}{{4 \cancel{\sqrt x} \sqrt {\sqrt x } }} = \frac{3}{{4\sqrt[4]{x}}}\;\;\left( {x \gt 0} \right).\]
Example 42.
\[y = \sqrt \frac{{x - 1}}{{x + 1}}\]
Solution.
Applying the chain rule and the quotient rule, we get
\[y^\prime = \left( {\sqrt {\frac{{x - 1}}{{x + 1}}} } \right)^\prime = \frac{1}{{2\sqrt {\frac{{x - 1}}{{x + 1}}} }} \cdot \left( {\frac{{x - 1}}{{x + 1}}} \right)^\prime = \frac{{2\sqrt {x + 1} }}{{\sqrt {x - 1} }} \cdot \frac{{1 \cdot \left( {x + 1} \right) - \left( {x - 1} \right) \cdot 1}}{{{{\left( {x + 1} \right)}^2}}} = \frac{{2\sqrt {x + 1} }}{{\sqrt {x - 1} }} \cdot \frac{{\cancel{x} + 1 - \cancel{x} + 1}}{{{{\left( {x + 1} \right)}^2}}} = \frac{{4\sqrt {x + 1} }}{{\sqrt {x - 1} {{\left( {x + 1} \right)}^2}}} = \frac{{4\sqrt {x + 1} }}{{\sqrt {\left( {x - 1} \right)\left( {x + 1} \right)} \sqrt {{{\left( {x + 1} \right)}^3}} }} = \frac{4}{{\sqrt {\left( {{x^2} - 1} \right)} \left( {x + 1} \right)}}.\]
Example 43.
\[y = {\sin ^4}x + {\cos ^4}x\]
Solution.
\[y'\left( x \right) = \left( {{{\sin }^4}x + {{\cos }^4}x} \right)^\prime = {\left( {{{\sin }^4}x} \right)^\prime } + {\left( {{{\cos }^4}x} \right)^\prime } = 4\,{\sin ^3}x \cdot {\left( {\sin x} \right)^\prime } + 4\,{\cos ^3}x \cdot {\left( {\cos x} \right)^\prime } = 4\,{\sin ^3}x\cos x - 4\,{\cos ^3}x\sin x = 4\sin x\cos x \cdot \left( {{{\sin }^2}x - {{\cos }^2}x} \right).\]
Next, using the double angle formula, we obtain a simple answer:
\[y'\left( x \right) = 4\sin x\cos x \cdot \left( {{{\sin }^2}x - {{\cos }^2}x} \right) = - 2 \cdot 2\sin x\cos x \cdot \left( {{\cos^2}x - {\sin^2}x} \right) = - 2\sin 2x\cos 2x = - \sin 4x.\]
Example 44.
\[y = \arccos \frac{a}{x}\]
Solution.
This function is defined for \(x \in \left( { - \infty , - a} \right] \cup \left[ {a,\infty } \right).\) Taking into account that
\[\left( {\arccos x} \right)^\prime = - \frac{1}{{\sqrt {1 - {x^2}} }},\]
and using the chain rule, we have
\[y'\left( x \right) = \left( {\arccos \frac{a}{x}} \right)^\prime = - \frac{1}{{\sqrt {1 - {{\left( {\frac{a}{x}} \right)}^2}} }} \cdot {\left( {\frac{a}{x}} \right)^\prime } = - \frac{1}{{\sqrt {1 - \frac{{{a^2}}}{{{x^2}}}} }} \cdot \left( { - \frac{a}{{{x^2}}}} \right) = \frac{1}{{\sqrt {\frac{{{x^2} - {a^2}}}{{{x^2}}}} }} \cdot \frac{a}{{{x^2}}} = \frac{{\left| x \right|}}{{\sqrt {{x^2} - {a^2}} }} \cdot \frac{a}{{{{\left| x \right|}^2}}} = \frac{a}{{\left| x \right|\sqrt {{x^2} - {a^2}} }}.\]
Example 45.
\[y = \text{arccot}\,\frac{{{x^2}}}{a}\;\left( {a \ne 0} \right)\]
Solution.
We know the derivative of the inverse tangent function:
\[\left( {\text{arccot}\,x} \right)^\prime = - \frac{1}{{1 + {x^2}}}.\]
Then differentiating as a composite function, we find:
\[y'\left( x \right) = \left( {\text{arccot}\,\frac{{{x^2}}}{a}} \right)^\prime = - \frac{1}{{1 + {{\left( {\frac{{{x^2}}}{a}} \right)}^2}}} \cdot {\left( {\frac{{{x^2}}}{a}} \right)^\prime } = - \frac{1}{{1 + \frac{{{x^4}}}{{{a^2}}}}} \cdot \frac{{2x}}{a} = - \frac{1}{{\frac{{{a^2} + {x^4}}}{{{a^2}}}}} \cdot \frac{{2x}}{a} = - \frac{{2{a^2}x}}{{a\left( {{a^2} + {x^4}} \right)}} = - \frac{{2ax}}{{{a^2} + {x^4}}}.\]
Example 46.
\[y = \sqrt[3]{{{{\cot }^8}\frac{x}{2}}}\]
Solution.
Applying the chain rule twice, we have
\[y'\left( x \right) = \left( {\sqrt[3]{{{{\cot }^8}\frac{x}{2}}}} \right)^\prime = {\left[ {{{\left( {\cot \frac{x}{2}} \right)}^{\frac{8}{3}}}} \right]^\prime } = \frac{8}{3}{\left( {\cot \frac{x}{2}} \right)^{\frac{8}{3} - 1}} \cdot {\left( {\cot \frac{x}{2}} \right)^\prime } = \frac{8}{3}{\left( {\cot \frac{x}{2}} \right)^{\frac{5}{3}}} \cdot \left( { - \frac{1}{{{{\sin }^2}\frac{x}{2}}}} \right) \cdot {\left( {\frac{x}{2}} \right)^\prime } = - \frac{8}{3}{\left( {\cot \frac{x}{2}} \right)^{\frac{5}{3}}} \cdot {\sec ^2}\frac{x}{2} \cdot \frac{1}{2} = - \frac{4}{3}\sqrt[3]{{{{\cot }^5}\frac{x}{2}}}{\sec ^2}\frac{x}{2}.\]
This function and its derivative exist when \(x \ne \pi n,\;n \in \mathbb{Z}.\)
Example 47.
\[y = {\log _5}\sin 2x\]
Solution.
The domain of the function is defined by the following condition:
\[\sin 2x \gt 0,\;\; \Rightarrow 2\pi n \lt 2x \lt \pi + 2\pi n\;\; \Rightarrow \text{or}\;\; \Rightarrow \pi n \lt x \lt \frac{\pi }{2} + \pi n,\;\;n \in \mathbb{Z}.\]
Differentating this function as a "three-layer" composite function, we have
\[y'\left( x \right) = \left( {{{\log }_5}\sin 2x} \right)^\prime = \frac{1}{{\ln 5\sin 2x}} \cdot {\left( {\sin 2x} \right)^\prime } = \frac{1}{{\ln 5\sin 2x}} \cdot \cos 2x \cdot 2 = \frac{{2\cos 2x}}{{\ln 5\sin 2x}} = \frac{{2\cot 2x}}{{\ln 5}}.\]
Example 48.
\[y = x\sin \frac{1}{x}\]
Solution.
First we find the derivative of the product:
\[y'\left( x \right) = \left( {x\sin \frac{1}{x}} \right)^\prime = {\left( x \right)^\prime }\sin \frac{1}{x} + x{\left( {\sin \frac{1}{x}} \right)^\prime }.\]
Then, by the chain rule, we obtain:
\[y'\left( x \right) = 1 \cdot \sin \frac{1}{x} + x \cdot \cos \frac{1}{x} \cdot {\left( {\frac{1}{x}} \right)^\prime } = \sin \frac{1}{x} + x\cos \frac{1}{x} \cdot \left( { - \frac{1}{{{x^2}}}} \right) = \sin \frac{1}{x} - \frac{1}{x}\cos \frac{1}{x}.\]
Example 49.
\[y = \ln \left( {x + \sqrt {{x^2} + 1} } \right)\]
Solution.
Differentiating the argument of the logarithm as a composite function, we have
\[y'\left( x \right) = \left[ {\ln \left( {x + \sqrt {{x^2} + 1} } \right)} \right]^\prime = \frac{1}{{x + \sqrt {{x^2} + 1} }} \cdot {\left( {x + \sqrt {{x^2} + 1} } \right)^\prime } = \frac{{1 + \frac{1}{{2\sqrt {{x^2} + 1} }} \cdot {{\left( {{x^2} + 1} \right)}^\prime }}}{{x + \sqrt {{x^2} + 1} }} = \frac{{1 + \frac{{\cancel{2}x}}{{\cancel{2}\sqrt {{x^2} + 1} }}}}{{x + \sqrt {{x^2} + 1} }} = \frac{{\frac{{\sqrt {{x^2} + 1} + x}}{{\sqrt {{x^2} + 1} }}}}{{x + \sqrt {{x^2} + 1} }} = \frac{\cancel{x + \sqrt {{x^2} + 1} }}{{\sqrt {{x^2} + 1} \cancel{\left( {x + \sqrt {{x^2} + 1} } \right)}}} = \frac{1}{{\sqrt {{x^2} + 1} }}.\]
One can see that in this case the function and the derivative exist for any real \(x.\)
Example 50.
\[y = \ln \frac{1}{{\sqrt {1 - {x^4}} }}\]
Solution.
We apply the quotient rule, the chain rule and the power rule to find the derivative.
\[y^\prime = \left( {\ln \frac{1}{{\sqrt {1 - {x^4}} }}} \right)^\prime = \frac{1}{{\frac{1}{{\sqrt {1 - {x^4}} }}}} \cdot \left( {\frac{1}{{\sqrt {1 - {x^4}} }}} \right)^\prime = \sqrt {1 - {x^4}} \cdot \left( {{{\left( {1 - {x^4}} \right)}^{ - \frac{1}{2}}}} \right)^\prime = \sqrt {1 - {x^4}} \cdot \left( { - \frac{1}{2}{{\left( {1 - {x^4}} \right)}^{ - \frac{3}{2}}}} \right) \cdot \left( {1 - {x^4}} \right)^\prime = \sqrt {1 - {x^4}} \cdot \left( { - \frac{1}{{2\sqrt {{{\left( {1 - {x^4}} \right)}^3}} }}} \right) \cdot \left( { - 4{x^3}} \right) = \frac{{2{x^3}\sqrt {1 - {x^4}} }}{{\sqrt {{{\left( {1 - {x^4}} \right)}^3}} }} = \frac{{2{x^3}}}{{1 - {x^4}}}.\]
Example 51.
\[y = \sin \left[ {\sin\left( {\sin x} \right)} \right]\]
Solution.
Here again we are dealing with a "three-layer" function. Therefore, we apply the chain rule twice. As a result, we obtain the following expression for the derivative:
\[y'\left( x \right) = \left( {\sin \left[ {\sin\left( {\sin x} \right)} \right]} \right)^\prime = \cos \left[ {\sin\left( {\sin x} \right)} \right] \cdot {\left[ {\sin\left( {\sin x} \right)} \right]^\prime } = \cos \left[ {\sin\left( {\sin x} \right)} \right] \cdot \cos\left( {\sin x} \right) \cdot {\left( {\sin x} \right)^\prime } = \cos \left[ {\sin\left( {\sin x} \right)} \right] \cos\left( {\sin x} \right) \cos x.\]
Example 52.
\[y = \frac{1}{{{{\cos }^n}x}}\]
Solution.
It is obvious that here \(x \ne {\frac{\pi }{2}} + \pi n,\) \(n \in \mathbb{Z}.\) Using the quotient rule and the chain rule, we get:
\[y'\left( x \right) = \left( {\frac{1}{{{{\cos }^n}x}}} \right)^\prime = \frac{{1' \cdot {{\cos }^n}x - 1 \cdot {{\left( {{{\cos }^n}x} \right)}^\prime }}}{{{{\left( {{{\cos }^n}x} \right)}^2}}} = \frac{{ - n\,{{\cos }^{n - 1}}x \cdot \left( { - \sin x} \right)}}{{{{\cos }^{2n}}x}} = \frac{{n\,{{\cos }^n}x \cdot {{\cos }^{ - 1}}x\sin x}}{{{{\cos }^{2n}}x}} = \frac{{n\sin x}}{{{{\cos }^n}x\cos x}} = \frac{{n\tan x}}{{{{\cos }^n}x}}.\]
Example 53.
\[y = \frac{2}{3}\sqrt {{{\left( {1 + \ln x} \right)}^3}} \]
Solution.
Differentiating twice as a composite function, we have
\[y'\left( x \right) = {\left[ {\frac{2}{3}\sqrt {{{\left( {1 + \ln x} \right)}^3}} } \right]^\prime } = \frac{2}{3} \cdot \frac{1}{{2\sqrt {{{\left( {1 + \ln x} \right)}^3}} }} \cdot {\left[ {{{\left( {1 + \ln x} \right)}^3}} \right]^\prime } = \frac{1}{{3\sqrt {{{\left( {1 + \ln x} \right)}^3}} }} \cdot 3{\left( {1 + \ln x} \right)^2} \cdot {\left( {1 + \ln x} \right)^\prime } = \frac{{\cancel{3}{{\left( {1 + \ln x} \right)}^2}}}{{\cancel{3}{{\left( {1 + \ln x} \right)}^{\frac{3}{2}}}}} \cdot \frac{1}{x} = \frac{{{{\left( {1 + \ln x} \right)}^{\frac{1}{2}}}}}{x} = \frac{{\sqrt {1 + \ln x} }}{x}.\]
The domain of the function and the derivative satisfies the condition:
\[\left\{ \begin{array}{l} 1 + \ln x \ge 0\\ x \gt 0 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} \ln x \ge – 1\\ x \gt 0 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} \ln x \ge \ln \frac{1}{e}\\ x \gt 0 \end{array} \right.,\;\; \Rightarrow x \ge \frac{1}{e}.\]
Example 54.
\[y = \ln \frac{1}{{\sqrt {1 - {x^4}} }}\]
Solution.
Here we encounter with a "four-layer" composite function. Its derivative is given by
\[y'\left( x \right) = \left( {\ln \frac{1}{{\sqrt {1 - {x^4}} }}} \right)^\prime = \frac{1}{{\frac{1}{{\sqrt {1 - {x^4}} }}}} \cdot {\left( {\frac{1}{{\sqrt {1 - {x^4}} }}} \right)^\prime } = \sqrt {1 - {x^4}} \cdot \frac{{1' \cdot \sqrt {1 - {x^4}} - 1 \cdot {{\left( {\sqrt {1 - {x^4}} } \right)}^\prime }}}{{{{\left( {\sqrt {1 - {x^4}} } \right)}^2}}} = \sqrt {1 - {x^4}} \cdot \frac{{0 - \frac{1}{{2\sqrt {1 - {x^4}} }} \cdot {{\left( {1 - {x^4}} \right)}^\prime }}}{{\left( {1 - {x^4}} \right)}} = \sqrt {1 - {x^4}} \cdot \frac{{ - \left( { - 4{x^3}} \right)}}{{2\sqrt {1 - {x^4}} \left( {1 - {x^4}} \right)}} = \frac{{4{x^3}\cancel{\sqrt {1 - {x^4}} }}}{{2\cancel{\sqrt {1 - {x^4}}} \left( {1 - {x^4}} \right)}} = \frac{{2{x^3}}}{{1 - {x^4}}}.\]
The function and the derivative are defined for \( - 1 \lt x \lt 1.\)
Example 55.
\[y = \ln \tan \left( {\frac{x}{2} + \frac{\pi }{4}} \right)\]
Solution.
This function can be considered as a "three-layer" composite function. Therefore its derivative is written as
\[y'\left( x \right) = \left[ {\ln \tan \left( {\frac{x}{2} + \frac{\pi }{4}} \right)} \right]^\prime = \frac{1}{{\tan \left( {\frac{x}{2} + \frac{\pi }{4}} \right)}} \cdot {\left[ {\tan \left( {\frac{x}{2} + \frac{\pi }{4}} \right)} \right]^\prime } = \frac{1}{{\tan \left( {\frac{x}{2} + \frac{\pi }{4}} \right)}} \cdot \frac{1}{{{\cos^2}\left( {\frac{x}{2} + \frac{\pi }{4}} \right)}} \cdot {\left( {\frac{x}{2} + \frac{\pi }{4}} \right)^\prime } = \frac{{\cos\left( {\frac{x}{2} + \frac{\pi }{4}} \right)}}{{\sin\left( {\frac{x}{2} + \frac{\pi }{4}} \right)}} \cdot \frac{1}{{{\cos^2}\left( {\frac{x}{2} + \frac{\pi }{4}} \right)}} \cdot \frac{1}{2} = \frac{1}{{2\sin\left( {\frac{x}{2} + \frac{\pi }{4}} \right) \cos\left( {\frac{x}{2} + \frac{\pi }{4}} \right)}}.\]
We simplify this expression by the double angle formula:
\[y'\left( x \right) = \frac{1}{{2\sin\left( {\frac{x}{2} + \frac{\pi }{4}} \right)\cos\left( {\frac{x}{2} + \frac{\pi }{4}} \right)}} = \frac{1}{{\sin\left( {2\left( {\frac{x}{2} + \frac{\pi }{4}} \right)} \right)}} = \frac{1}{{\sin\left( {x + \frac{\pi }{2}} \right)}}.\]
Finally, using the reduction identity, we get the final answer:
\[y'\left( x \right) = \frac{1}{{\sin\left( {x + \frac{\pi }{2}} \right)}} = \frac{1}{{\cos x}} = \csc x.\]
Find the domain of the function. It is defined by the condition
\[\tan \left( {\frac{x}{2} + \frac{\pi }{4}} \right) \gt 0.\]
Hence,
\[\pi n \lt \frac{x}{2} + \frac{\pi }{4} \lt \frac{\pi }{2} + \pi n,\;\; \Rightarrow - \frac{\pi }{4} + \pi n \lt \frac{x}{2} \lt \frac{\pi }{4} + \pi n,\;\; \Rightarrow - \frac{\pi }{2} + 2\pi n \lt x \lt \frac{\pi }{2} + 2\pi n,\;\text{where}\;\;n \in \mathbb{Z}.\]
