# 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:

$\left\{ \begin{array}{l} \sin x \gt 0\\ \sin x \ne 1 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} \pi n \lt x \lt \pi + \pi n,\;n \in Z\\ x \ne \frac{\pi }{2} + \pi k,\;k \in Z \end{array} \right.\;\; \Rightarrow {\text{or}\;\;x \in \left( {\pi n,\frac{\pi }{2} + \pi n} \right) } \cup { \left( {\frac{\pi }{2} + \pi n,\pi + \pi n} \right),\;n \in Z.}$