Logarithmic Differentiation
Solved Problems
Example 9.
\[y = {\frac{{{{\left( {x - 2} \right)}^2}}}{{{{\left( {x + 5} \right)}^3}}}},\] where \(x \gt 2.\)
Solution.
Take logarithms of both sides:
\[\ln y = \ln \frac{{{{\left( {x - 2} \right)}^2}}}{{{{\left( {x + 5} \right)}^3}}},\;\; \Rightarrow \ln y = \ln {\left( {x - 2} \right)^2} - \ln {\left( {x + 5} \right)^3}, \Rightarrow \ln y = 2\ln \left( {x - 2} \right) - 3\ln \left( {x + 5} \right).\]
Find the logarithmic derivative:
\[\require{cancel} \left( {\ln y} \right)^\prime = \left( {2\ln \left( {x - 2} \right)} \right)^\prime - \left( {3\ln \left( {x + 5} \right)} \right)^\prime,\;\; \Rightarrow \frac{{y^\prime}}{y} = \frac{2}{{x - 2}} - \frac{3}{{x + 5}},\;\; \Rightarrow \frac{{y^\prime}}{y} = \frac{{2\left( {x + 5} \right) - 3\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 5} \right)}},\;\; \Rightarrow \frac{{y^\prime}}{y} = \frac{{\color{blue}{2x} + \color{red}{10} - \color{blue}{3x} + \color{red}{6}}}{{\left( {x - 2} \right)\left( {x + 5} \right)}},\;\; \Rightarrow \frac{{y^\prime}}{y} = \frac{{\color{red}{16} - \color{blue}{x}}}{{\left( {x - 2} \right)\left( {x + 5} \right)}},\;\; \Rightarrow y^\prime = y\frac{{16 - x}}{{\left( {x - 2} \right)\left( {x + 5} \right)}},\;\; \Rightarrow y^\prime = \frac{{{{\left( {x - 2} \right)}^\cancel{2}}\left( {16 - x} \right)}}{{{{\left( {x + 5} \right)}^3}\cancel{\left( {x - 2} \right)}\left( {x + 5} \right)}},\;\; \Rightarrow y^\prime = \frac{{\left( {x - 2} \right)\left( {16 - x} \right)}}{{{{\left( {x + 5} \right)}^4}}}.\]
Example 10.
\[y\left( x \right) = \frac{{{{\left( {x + 1} \right)}^2}}}{{{{\left( {x + 2} \right)}^3}{{\left( {x + 3} \right)}^4}}},\] where \(x \gt - 1.\)
Solution.
Take logarithms of both sides:
\[\ln y = \ln \frac{{{{\left( {x + 1} \right)}^2}}}{{{{\left( {x + 2} \right)}^3}{{\left( {x + 3} \right)}^4}}},\;\; \Rightarrow \ln y = \ln {\left( {x + 1} \right)^2} - \ln {\left( {x + 2} \right)^3} - \ln {\left( {x + 3} \right)^4},\;\; \Rightarrow \ln y = 2\ln \left( {x + 1} \right) - 3\ln \left( {x + 2} \right) - 4\ln \left( {x + 3} \right).\]
Now differentiate the left and right sides:
\[\frac{{y'}}{y} = \frac{2}{{x + 1}} - \frac{3}{{x + 2}} - \frac{4}{{x + 3}},\;\; \Rightarrow y' = y \cdot \left( {\frac{2}{{x + 1}} - \frac{3}{{x + 2}} - \frac{4}{{x + 3}}} \right),\;\; \Rightarrow y' = \frac{{{{\left( {x + 1} \right)}^2}}}{{{{\left( {x + 2} \right)}^3}{{\left( {x + 3} \right)}^4}}} \cdot \left( {\frac{2}{{x + 1}} - \frac{3}{{x + 2}} - \frac{4}{{x + 3}}} \right).\]
Example 11.
\[y = \sqrt[x]{x},\;x \gt 0.\]
Solution.
Take the natural logarithm of this equation:
\[\ln y = \ln \left( {\sqrt[x]{x}} \right),\;\; \Rightarrow \ln y = \frac{1}{x}\ln x.\]
Differentiating both sides with respect to \(x\) yields:
\[{\left( {\ln y} \right)^\prime } = {\left( {\frac{1}{x}\ln x} \right)^\prime },\;\; \Rightarrow {\frac{1}{y} \cdot y' = {\left( {\frac{1}{x}} \right)^\prime }\ln x + \frac{1}{x}{\left( {\ln x} \right)^\prime },\;\;} \Rightarrow {\frac{{y'}}{y} = \left( { - \frac{1}{{{x^2}}}} \right) \cdot \ln x + \frac{1}{x} \cdot \frac{1}{x},\;\;} \Rightarrow {\frac{{y'}}{y} = \frac{1}{{{x^2}}}\left( {1 - \ln x} \right),\;\;} \Rightarrow y' = \frac{y}{{{x^2}}}\left( {1 - \ln x} \right),\;\; \Rightarrow y' = \frac{{\sqrt[x]{x}}}{{{x^2}}}\left( {1 - \ln x} \right)\]
or
\[y' = {x^{\frac{1}{x} - 2}}\left( {1 - \ln x} \right),\;\;\text{where}\;\;x \gt 0.\]
Example 12.
\[y = {\left( {\ln x} \right)^x},\;x \gt 1.\]
Solution.
Following the common pattern, we can write:
\[\ln y = \ln \left[ {{{\left( {\ln x} \right)}^x}} \right],\;\; \Rightarrow \ln y = x\ln \left( {\ln x} \right).\]
Calculate the logarithmic derivative:
\[{\left( {\ln y} \right)^\prime } = {\left[ {x\ln \left( {\ln x} \right)} \right]^\prime },\;\; \Rightarrow \frac{1}{y} \cdot y' = x' \cdot \ln \left( {\ln x} \right) + x \cdot {\left[ {\ln \left( {\ln x} \right)} \right]^\prime },\;\; \Rightarrow \frac{{y'}}{y} = 1 \cdot \ln \left( {\ln x} \right) + x \cdot \frac{1}{{\ln x}} \cdot {\left( {\ln x} \right)^\prime },\;\; \Rightarrow \frac{{y'}}{y} = \ln \left( {\ln x} \right) + x \cdot \frac{1}{{\ln x}} \cdot \frac{1}{x},\;\; \Rightarrow \frac{{y'}}{y} = \ln \left( {\ln x} \right) + \frac{1}{{\ln x}},\;\; \Rightarrow y' = y\left[ {\ln \left( {\ln x} \right) + \frac{1}{{\ln x}}} \right],\;\; \Rightarrow y' = {\left( {\ln x} \right)^x}\left[ {\ln \left( {\ln x} \right) + \frac{1}{{\ln x}}} \right].\]
Example 13.
\[y = {\left( {{e^x}} \right)^{{e^x}}}.\]
Solution.
Taking logarithms of both sides, we have
\[\ln y = \ln {\left( {{e^x}} \right)^{{e^x}}},\;\; \Rightarrow \ln y = {e^x}\ln \left( {{e^x}} \right),\;\; \Rightarrow \ln y = {e^x}x.\]
Differentiate this expression with respect to \(x:\)
\[\left( {\ln y} \right)^\prime = \left( {{e^x}x} \right)^\prime,\;\; \Rightarrow \frac{1}{y} \cdot y^\prime = \left( {{e^x}} \right)^\prime \cdot x + {e^x} \cdot x^\prime,\;\; \Rightarrow \frac{{y^\prime}}{y} = {e^x} \cdot x + {e^x} \cdot 1,\;\; \Rightarrow \frac{{y^\prime}}{y} = {e^x}\left( {x + 1} \right).\]
Multiply both sides by \(y\) and simplify:
\[y^\prime = y{e^x}\left( {x + 1} \right) = {\left( {{e^x}} \right)^{{e^x}}}{e^x}\left( {x + 1} \right) = {\left( {{e^x}} \right)^{{e^x} + 1}}\left( {x + 1} \right).\]
Example 14.
\[y = {\left( {\ln x} \right)^{\ln x}},\;x \gt 1.\]
Solution.
Following the general approach, we can write:
\[\ln y = \ln \left[ {{{\left( {\ln x} \right)}^{\ln x}}} \right],\;\; \Rightarrow \ln y = \ln x \cdot \ln \left( {\ln x} \right).\]
Differentiate both sides with respect to \(x:\)
\[\left( {\ln y} \right)^\prime = \left[ {\ln x \cdot \ln \left( {\ln x} \right)} \right]^\prime,\;\; \Rightarrow \frac{1}{y} \cdot y^\prime = \left( {\ln x} \right)^\prime \cdot \ln \left( {\ln x} \right) + \ln x \cdot \left( {\ln \left( {\ln x} \right)} \right)^\prime,\;\; \Rightarrow \frac{{y^\prime}}{y} = \frac{1}{x} \cdot \ln \left( {\ln x} \right) + \ln x \cdot \frac{1}{{\ln x}} \cdot \frac{1}{x},\;\; \Rightarrow \frac{{y^\prime}}{y} = \frac{{\ln \left( {\ln x} \right)}}{x} + \frac{1}{x},\;\; \Rightarrow y^\prime = \frac{y}{x}\left[ {\ln \left( {\ln x} \right) + 1} \right],\;\; \Rightarrow y^\prime = \frac{{{{\left( {\ln x} \right)}^{\ln x}}}}{x}\left[ {\ln \left( {\ln x} \right) + 1} \right].\]
Example 15.
\[y = {x^{{x^2}}}\;\left( {x \gt 0,x \ne 1} \right)\]
Solution.
Taking logarithms of both sides yields:
\[\ln y = \ln \left( {{x^{{x^2}}}} \right),\;\; \Rightarrow \ln y = {x^2}\ln x.\]
Now it is easy to calculate the derivative:
\[{\left( {\ln y} \right)^\prime } = {\left( {{x^2}\ln x} \right)^\prime },\;\; \Rightarrow {\frac{1}{y} \cdot y' = {\left( {{x^2}} \right)^\prime } \cdot \ln x + {x^2} \cdot {\left( {\ln x} \right)^\prime },\;\;} \Rightarrow {\frac{{y'}}{y} = 2x \cdot \ln x + {x^2} \cdot \frac{1}{x},\;\;} \Rightarrow \frac{{y'}}{y} = 2x\ln x + x,\;\; \Rightarrow y' = yx\left( {2\ln x + 1} \right),\;\; \Rightarrow y' = {x^{{x^2}}}x\left( {2\ln x + 1} \right)\;\; \Rightarrow \text{or}\;\;y' = {x^{{x^2} + 1}}\left( {2\ln x + 1} \right).\]
Example 16.
\[y = {x^{{x^n}}}\;\left( {x \gt 0, x \ne 1} \right)\]
Solution.
Take logarithms of both sides:
\[\ln y = \ln \left( {{x^{{x^n}}}} \right),\;\; \Rightarrow \;\ln y = {x^n}\ln x.\]
Hence,
\[{\left( {\ln y} \right)^\prime } = {\left( {{x^n}\ln x} \right)^\prime },\;\; \Rightarrow {\frac{1}{y} \cdot y' = {\left( {{x^n}} \right)^\prime } \cdot \ln x + {x^n} \cdot {\left( {\ln x} \right)^\prime },\;\;} \Rightarrow {\frac{{y'}}{y} = n{x^{n - 1}} \cdot \ln x + {x^n} \cdot \frac{1}{x},\;\;} \Rightarrow {\frac{{y'}}{y} = n{x^{n - 1}}\ln x + {x^{n - 1}},\;\;} \Rightarrow y' = y{x^{n - 1}}\left( {n\ln x + 1} \right),\;\; \Rightarrow y' = {x^{{x^n}}}{x^{n - 1}}\left( {n\ln x + 1} \right)\;\; \Rightarrow y' = {x^{{x^n} + n - 1}}\left( {n\ln x + 1} \right).\]
Example 17.
\[y = {x^{{2^x}}}\;\left( {x \gt 0, x \ne 1} \right)\]
Solution.
\[\ln y = \ln \left( {{x^{{2^x}}}} \right),\;\; \Rightarrow \ln y = {2^x}\ln x.\]
Differentiate both sides of the equality:
\[{\left( {\ln y} \right)^\prime } = {\left( {{2^x}\ln x} \right)^\prime },\;\; \Rightarrow {\frac{1}{y} \cdot y' = {\left( {{2^x}} \right)^\prime } \cdot \ln x + {2^x} \cdot {\left( {\ln x} \right)^\prime },\;\;} \Rightarrow {\frac{{y'}}{y} = {2^x}\ln 2 \cdot \ln x + {2^x} \cdot \frac{1}{x},\;\;} \Rightarrow {\frac{{y'}}{y} = {2^x}\left( {\ln 2\ln x + \frac{1}{x}} \right),\;\;} \Rightarrow {y' = {x^{{2^x}}}{2^x}\left( {\ln 2\ln x + \frac{1}{x}} \right).}\]
Example 18.
\[y = {2^{{x^x}}}\;\left( {x \gt 0, x \ne 1} \right)\]
Solution.
Using the chain rule, we find:
\[y' = {\left( {{2^{{x^x}}}} \right)^\prime } = {2^{{x^x}}} \cdot {\left( {{x^x}} \right)^\prime }.\]
The derivative of the power-exponential function \({{x^x}}\) was determined in Example \(1.\) Substituting it, we have:
\[y' = {\left( {{2^{{x^x}}}} \right)^\prime } = {2^{{x^x}}} \cdot {\left( {{x^x}} \right)^\prime } = {2^{{x^x}}}{x^x}\left( {\ln x + 1} \right).\]
Example 19.
\[y = {x^{\sqrt x }}\;\left( {x \gt 0, x \ne 1} \right)\]
\[\ln y = \ln \left( {{x^{\sqrt x }}} \right),\;\; \Rightarrow \ln y = \sqrt x \ln x.\]
We find from here:
\[{\left( {\ln y} \right)^\prime } = {\left( {\sqrt x \ln x} \right)^\prime },\;\; \Rightarrow \frac{1}{y} \cdot y' = {\left( {\sqrt x } \right)^\prime } \cdot \ln x + \sqrt x \cdot {\left( {\ln x} \right)^\prime },\;\; \Rightarrow {\frac{{y'}}{y} = \frac{1}{{2\sqrt x }} \cdot \ln x + \sqrt x \cdot \frac{1}{x},\;\;} \Rightarrow {\frac{{y'}}{y} = \frac{1}{{2\sqrt x }}\left( {\ln x + 2} \right),\;\;} \Rightarrow y' = \frac{y}{{2\sqrt x }}\left( {\ln x + 2} \right),\;\; \Rightarrow y' = \frac{{{x^{\sqrt x }}}}{{2\sqrt x }}\left( {\ln x + 2} \right).\]
Example 20.
\[y = {x^{{x^x}}}\;\left( {x \gt 0, x \ne 1} \right)\]
Solution.
Take the logarithm of the given function:
\[\ln y = \ln \left( {{x^{{x^x}}}} \right),\;\; \Rightarrow \ln y = {x^x}\ln x.\]
Differentiating both sides with respect to \(x,\) we obtain:
\[\left( {\ln y} \right)^\prime = \left( {{x^x}\ln x} \right)^\prime ,\;\;\Rightarrow {\frac{1}{y} \cdot y' = {\left( {{x^x}} \right)^\prime } \cdot \ln x + {x^x} \cdot {\left( {\ln x} \right)^\prime },\;\;} \Rightarrow {\frac{{y'}}{y} = {\left( {{x^x}} \right)^\prime } \cdot \ln x + {x^x} \cdot \frac{1}{x}.}\]
In Example \(1,\) we have calculated the derivative of the function \(y = {x^x}\). Substituting it into the upper relation, we obtain the following expression for the derivative of the original function:
\[\frac{{y'}}{y} = {x^x}\left( {\ln x + 1} \right) \cdot \ln x + {x^x} \cdot \frac{1}{x},\;\; \Rightarrow \frac{{y'}}{y} = {x^x}\cdot \left[ {\ln x\left( {\ln x + 1} \right) + \frac{1}{x}} \right],\;\; \Rightarrow {y' = y{x^x}\cdot \left[ {\ln x\left( {\ln x + 1} \right) + \frac{1}{x}} \right],\;\;} \Rightarrow {y' = {x^{{x^x}}}{x^x}\cdot \left[ {\ln x\left( {\ln x + 1} \right) + \frac{1}{x}} \right],\;\;} \Rightarrow {y' = {x^{{x^x} + x}} \left[ {\ln x\left( {\ln x + 1} \right) + \frac{1}{x}} \right].} \]
Example 21.
\[y = {\sqrt x ^{\sqrt x }}\;\left( {x \gt 0, x \ne 1} \right)\]
Solution.
\[\ln y = \ln \left( {{{\sqrt x }^{\sqrt x }}} \right),\;\; \Rightarrow \ln y = \sqrt x \ln \sqrt x .\]
Consequently,
\[{\left( {\ln y} \right)^\prime } = {\left( {\sqrt x \ln \sqrt x } \right)^\prime },\;\; \Rightarrow {\frac{1}{y} \cdot y' = {\left( {\sqrt x } \right)^\prime } \cdot \ln \sqrt x + \sqrt x \cdot {\left( {\ln \sqrt x } \right)^\prime },\;\;} \Rightarrow \frac{{y'}}{y} = \frac{1}{{2\sqrt x }} \cdot \ln \sqrt x + \sqrt x \cdot \frac{1}{{\sqrt x }} \cdot {\left( {\sqrt x } \right)^\prime },\;\; \Rightarrow \frac{{y'}}{y} = \frac{{\ln \sqrt x }}{{2\sqrt x }} + \frac{1}{{2\sqrt x }},\;\; \Rightarrow y' = \frac{y}{{2\sqrt x }}\left( {\ln \sqrt x + 1} \right),\;\; \Rightarrow y' = \frac{{{{\sqrt x }^{\sqrt x }}}}{{2\sqrt x }}\left( {\ln \sqrt x + 1} \right),\;\; \Rightarrow y' = \frac{1}{2}{\sqrt x ^{\sqrt x - 1}}\left( {\ln \sqrt x + 1} \right).\]
Example 22.
\[y = {\left( {\sin x} \right)^{\cos x}}\]
Solution.
It is assumed in this example that the variable \(x\) satisfies the domain constraints which have the form:
\[ \left\{ \begin{array}{l} \sin x \gt 0\\ \sin x \ne 1 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} 2\pi n \lt x \lt \pi + 2\pi n,\;{n \in \mathbb{Z}}\\ x \ne \frac{\pi }{2} + 2\pi m,\;m \in \mathbb{Z} \end{array} \right.,\;\; \Rightarrow {x \in \left( {2\pi n,\frac{\pi }{2} + 2\pi n} \right) } \cup { \left( {\frac{\pi }{2} + 2\pi n,\pi + 2\pi n} \right), n \in \mathbb{Z}.} \]
We take logarithms of both sides of the equality and then differentiate.
\[\ln y = \ln \left( {\sin {x^{\cos x}}} \right),\;\; \Rightarrow \ln y = \cos x\ln \sin x,\;\; \Rightarrow {\left( {\ln y} \right)^\prime } = {\left( {\cos x\ln \sin x} \right)^\prime },\;\; \Rightarrow {\frac{{y'}}{y} = {\left( {\cos x} \right)^\prime }\ln \sin x + \cos x{\left( {\ln \sin x} \right)^\prime },\;\;} \Rightarrow {\frac{{y'}}{y} = - \sin x \cdot \ln \sin x } + { \cos x \cdot \frac{1}{{\sin x}} \cdot \cos x,\;\;} \Rightarrow {\frac{{y'}}{y} = \frac{{{{\cos }^2}x}}{{\sin x}} - \sin x\ln \sin x,\;\;} \Rightarrow {y' = y\cdot \left( {\cot x\cos x - \sin x\ln \sin x} \right),\;\;} \Rightarrow {y' = \left( {\sin x} \right)^{\cos x}} {\left( {\cot x\cos x - \sin x\ln \sin x} \right).}\]
Example 23.
\[y = \sqrt[3]{{{\frac{{x - 2}}{{x + 2}}}}},\;x \gt 2.\]
Solution.
First we take logarithms of both sides:
\[\ln y = \ln \sqrt[3]{{\frac{{x - 2}}{{x + 2}}}},\;\; \Rightarrow \ln y = \frac{1}{3}\ln \left( {x - 2} \right) - \frac{1}{3}\ln \left( {x + 2} \right).\]
Differentiate this equation with respect to \(x:\)
\[\left( {\ln y} \right)^\prime = \left( {\frac{1}{3}\ln \left( {x - 2} \right)} \right)^\prime - \left( {\frac{1}{3}\ln \left( {x + 2} \right)} \right)^\prime,\;\; \Rightarrow \frac{{y^\prime}}{y} = \frac{1}{{3\left( {x - 2} \right)}} - \frac{1}{{3\left( {x + 2} \right)}},\;\; \Rightarrow \frac{{y^\prime}}{y} = \frac{{x + 2 - \left( {x - 2} \right)}}{{3\left( {x - 2} \right)\left( {x + 2} \right)}},\;\; \Rightarrow \frac{{y^\prime}}{y} = \frac{4}{{3\left( {x - 2} \right)\left( {x + 2} \right)}}.\]
Hence, the derivative is
\[y^\prime = y\frac{4}{{3\left( {x - 2} \right)\left( {x + 2} \right)}} = \sqrt[3]{{\frac{{x - 2}}{{x + 2}}}} \cdot \frac{4}{{3\left( {x - 2} \right)\left( {x + 2} \right)}} = \frac{{{{\left( {x - 2} \right)}^{\frac{1}{3}}}}}{{{{\left( {x + 2} \right)}^{\frac{1}{3}}}}} \cdot \frac{4}{{3\left( {x - 2} \right)\left( {x + 2} \right)}} = \frac{4}{{3{{\left( {x + 2} \right)}^{\frac{4}{3}}}{{\left( {x - 2} \right)}^{\frac{2}{3}}}}} = \frac{4}{{3\sqrt[3]{{{{\left( {x + 2} \right)}^4}{{\left( {x - 2} \right)}^2}}}}}.\]
Example 24.
\[y = \sqrt[3]{{\frac{{{x^2} - 3}}{{1 + {x^5}}}}},\;x \gt \sqrt 3 .\]
Solution.
Take logarithms of both sides:
\[\ln y = \ln \sqrt[3]{{\frac{{{x^2} - 3}}{{1 + {x^5}}}}},\;\; \Rightarrow \ln y = \ln {\left( {\frac{{{x^2} - 3}}{{1 + {x^5}}}} \right)^{\frac{1}{3}}} = {\frac{1}{3}\ln\left( {\frac{{{x^2} - 3}}{{1 + {x^5}}}} \right),}\;\; \Rightarrow {\ln y = \frac{1}{3}\left[ {\ln \left( {{x^2} - 3} \right) - \ln \left( {1 + {x^5}} \right)} \right].}\]
By differentiating the left and right side of the equation, we get:
\[{\left( {\ln y} \right)^\prime } = \frac{1}{3}{\left[ {\ln \left( {{x^2} - 3} \right) - \ln \left( {1 + {x^5}} \right)} \right]^\prime },\;\; \Rightarrow {\frac{{y'}}{y} = \frac{1}{3}\cdot} {\left[ {{{\left( {\ln \left( {{x^2} - 3} \right)} \right)}^\prime } - {{\left( {\ln \left( {1 + {x^5}} \right)} \right)}^\prime }} \right],\;\;} \Rightarrow {\frac{{y'}}{y} = \frac{1}{3}\cdot \left( {\frac{{2x}}{{{x^2} - 3}} - \frac{{5{x^4}}}{{1 + {x^5}}}} \right),\;\;} \Rightarrow {y' = \frac{y}{3}\cdot \left( {\frac{{2x}}{{{x^2} - 3}} - \frac{{5{x^4}}}{{1 + {x^5}}}} \right),\;\;} \Rightarrow {y' = \frac{1}{3}\sqrt[3]{{\frac{{{x^2} - 3}}{{1 + {x^5}}}}} \left( {\frac{{2x}}{{{x^2} - 3}} - \frac{{5{x^4}}}{{1 + {x^5}}}} \right).}\]
Example 25.
\[y = {\left( {\cos x} \right)^{\arcsin x}}\]
Solution.
Take logarithms of both sides of the equality:
\[\ln y = \ln \left[ {{{\left( {\cos x} \right)}^{\arcsin x}}} \right],\;\; \Rightarrow \ln y = \arcsin x\ln \cos x.\]
Then differentiating, we have
\[\left( {\ln y} \right)^\prime = \left( {\arcsin x\ln \cos x} \right)^\prime ,\;\; \Rightarrow \frac{1}{y} \cdot y' = {\left( {\arcsin x} \right)^\prime } \cdot \ln \cos x + \arcsin x \cdot {\left( {\ln \cos x} \right)^\prime },\;\; \Rightarrow \frac{{y'}}{y} = \frac{1}{{\sqrt {1 - {x^2}} }} \cdot \ln \cos x + \arcsin x \cdot \frac{1}{{\cos x}} \cdot {\left( {\cos x} \right)^\prime },\;\; \Rightarrow \frac{{y'}}{y} = \frac{{\ln \cos x}}{{\sqrt {1 - {x^2}} }} - \arcsin x\tan x,\;\; \Rightarrow y' = y \cdot \left( {\frac{{\ln \cos x}}{{\sqrt {1 - {x^2}} }} - \arcsin x\tan x} \right)\;\; \Rightarrow {\text{or}\;\;y' = {\left( {\cos x} \right)^{\arcsin x}}} {\left( {\frac{{\ln \cos x}}{{\sqrt {1 - {x^2}} }} - \arcsin x\tan x} \right).}\]
The domain of this function and its derivative is described by the following inequalities:
\[ \left\{ \begin{array}{l} \cos x \gt 0\\ \cos x \ne 1\\ \left| x \right| \lt 1 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} - \frac{\pi }{2} + 2\pi n \lt x \lt \frac{\pi }{2} + 2\pi n,\\ x \ne 2\pi k,\\ \left| x \right| \lt 1 \end{array} \right.\;\; \Rightarrow 0 \lt \left| x \right| \lt 1, \]
where \(n, k \in \mathbb{Z}.\)
Example 26.
\[y = {\left( {\sin x} \right)^{\arctan x}}\]
Solution.
\[\ln y = \ln \left[ {{{\left( {\sin x} \right)}^{\arctan x}}} \right],\;\; \Rightarrow \ln y = \arctan x\ln \sin x.\]
Differentiating both sides in \(x,\) we find the derivative:
\[\left( {\ln y} \right)^\prime = \left( {\arctan x\ln \sin x} \right)^\prime ,\;\; \Rightarrow \frac{1}{y} \cdot y' = {\left( {\arctan x} \right)^\prime } \cdot \ln \sin x + \arctan x \cdot {\left( {\ln \sin x} \right)^\prime },\;\; \Rightarrow \frac{{y'}}{y} = \frac{1}{{1 + {x^2}}} \cdot \ln \sin x + \arctan x \cdot \frac{1}{{\sin x}} \cdot {\left( {\sin x} \right)^\prime },\;\; \Rightarrow \frac{{y'}}{y} = \frac{{\ln \sin x}}{{1 + {x^2}}} + \arctan x\cot x,\;\; \Rightarrow y' = y \cdot \left( {\frac{{\ln \sin x}}{{1 + {x^2}}} + \arctan x\cot x} \right)\;\; \Rightarrow {\text{or}\;\;y' = {\left( {\sin x} \right)^{\arctan x}}} {\left( {\frac{{\ln \sin x}}{{1 + {x^2}}} + \arctan x\cot x} \right).}\]
Here the variable \(x\) satisfies the condition:
\[\left\{ \begin{array}{l} \sin x \gt 0\\ \sin x \ne 1 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} \pi n \lt x \lt \pi + \pi n,\;n \in Z\\ x \ne \frac{\pi }{2} + \pi k,\;k \in Z \end{array} \right.\;\; \Rightarrow {\text{or}\;\;x \in \left( {\pi n,\frac{\pi }{2} + \pi n} \right) } \cup { \left( {\frac{\pi }{2} + \pi n,\pi + \pi n} \right),\;n \in Z.} \]