The logarithmic derivation of a function \(y=f(x)\) is defined by the formula $$(\ln y)' = \frac{y'}{y} = \frac{f'(x)}{f(x)}.$$
The application of logarithmic function sometimes simplifies calculation of derivative. For eaxample let's calculate the derivative of a function $$y = u^v,$$ if \(u = \phi(x)>0\) and \(v = \psi(x)\) are differentiable. With application of logarithm $$\ln y = v \ln u. $$ With application of derivative on both sides withe respect to \(x\): \begin{eqnarray}(\ln y)' &=& v'\ln u + v (\ln u)'\\\nonumber \frac{1}{y}y' &=& v'\ln u + v \frac{1}{u}u'\\\nonumber y' &=& y\left(v' \ln u + \frac{v}{u}u'\right)\\\nonumber y' &=& u^v\left(v'\ln u + \frac{v}{u}u'\right)\end{eqnarray}
- Example 1 Calculate the \(y'\) of a function \(y = \sqrt[3]{x^2} \frac{1-x}{1+x^2}\sin^3 x\cos^2 x\)
Solution \begin{eqnarray}y &=& \sqrt[3]{x^2} \frac{1-x}{1+x^2}\sin^3 x\cos^2 x \\\nonumber \ln y &=& \frac{2}{3}\ln x + \ln(1-x) - \ln(1+x^2) + 3\ln \sin x + 2\ln \cos x \\\nonumber \frac{1}{y}y' &=& \frac{2}{3}\frac{1}{x}+ \frac{1}{1-x}(-1) - \frac{1}{1+x} + 3 \frac{1}{\sin x}\cos x+ 2 \frac{1}{\cos x}(-\sin x)\\\nonumber y' &=& \frac{(1 - x) x^{\frac{2}{3}} \cos^2 x \sin^3 x \left(\frac{2}{3x} + \frac{2}{x^2-1} + 3\cot x -2\tan x\right)}{(1 + x)}\end{eqnarray} - Example 2 Calculate the \(y'\) of a function \(y = (\sin x)^x\)
Solution \begin{eqnarray}\ln y &=& x\ln \sin x/' \\\nonumber \frac{1}{y} y' &=& \ln \sin x + x \frac{1}{\sin x} \cos x\\\nonumber \frac{1}{y}y' &=& \ln \sin x + x\cot x/\cdot y \\\nonumber y' &=& (\sin x)^x\left(\ln \sin x+ x\cot x\right)\end{eqnarray} - Example 3 Calculate the \(y'\) of a function \(y = (x+1)(2x+1)(3x+1)\)
Solution \begin{eqnarray}y &=& (x+1)(2x+1)(3x+1)/\ln \\\nonumber \ln y &=& \ln(x+1)(2x+1)(3x+1)\\\nonumber \ln y &=& \ln(x+1) + \ln(2x+1) + \ln(3x+1)/'\\\nonumber \frac{1}{y}y'&=& \frac{1}{x+1} + \frac{2}{2x+1}+\frac{3}{3x+1}\\\nonumber \frac{1}{y} y' &=& \frac{(2x+1)(3x+1) + 2(x+1)(3x+1) +3(x+1)(2x+1)}{(x+1)(2x+1)(3x+1)} \\\nonumber \frac{1}{y}y' &=& \frac{6x^2 + 2x+3x +1 + 2(3x^2 + x + 3x+1) + 3(2x^2 + x+ 2x+ 1)}{(x+1)(2x+1)(3x+1)}\\\nonumber \frac{1}{y}y'&=& \frac{18x^2 + 22x + 6}{(x+1)(2x+1)(3x+1)}/\cdot y\\\nonumber y' &=& y\frac{18x^2 + 22x + 6}{(x+1)(2x+1)(3x+1)}\\\nonumber y'&=& (x+1)(2x+1)(3x+1)\frac{18x^2 + 22x + 6}{(x+1)(2x+1)(3x+1)}\\\nonumber y'&=& 18x^2 + 22x + 6 \end{eqnarray} - Example 4 Calculate the \(y'\) of a function \(y = \frac{(x+2)^2}{(x+1)^3(x+3)^4}\)
Solution \begin{eqnarray}y &=& \frac{(x+2)^2}{(x+1)^3(x+3)^4}/\ln \\\nonumber \ln y &=& 2\ln(x+2) - \ln(x+1)^3(x+3)^4\\\nonumber \ln y &=& 2\ln(x+2) - 3\ln(x+1) - 4\ln(x+3)/'\\\nonumber \frac{1}{y}y'&=& \frac{2}{x+2} - \frac{3}{x+1} - \frac{4}{x+3} \\\nonumber \frac{1}{y}y'&=& \frac{2(x+1)(x+3) - 3(x+2)(x+3) - 3(x+1)(x+3)}{(x+1)(x+2)(x+3)} \\\nonumber \frac{1}{y}y'&=& \frac{2x^2 + 6x + 2x +6 - 3x^2 -9x-6x-18-4x^2 - 12x-12}{(x+1)(x+2)(x+3)}\\\nonumber \frac{1}{y}y'&=& \frac{-5x^2-23x-24}{(x+1)(x+2)(x+3)}/\cdot y\\\nonumber y'&=& - \frac{(x+2)^2}{(x+1)^3(x+3)^4} \frac{5x^2+23x+24}{(x+1)(x+2)(x+3)}\\\nonumber y' &=& -\frac{(x+2)(5x^2 + 23x + 24)}{(x+1)^4(x+3)^5}\end{eqnarray} - Example 5 Calculate the \(y'\) of a function \(y = \sqrt{\frac{x(x-1)}{x-2}}\)
Solution \begin{eqnarray}y &=& \sqrt{\frac{x(x-1)}{x-2}}/\ln \\\nonumber \ln y &=& \frac{1}{2}\frac{x(x-1)}{x-2} = \frac{1}{2}\ln x + \frac{1}{2}\ln(x-1) -\frac{1}{2}\ln(x-2)\\\nonumber \frac{1}{y}y' &=& \frac{1}{2}\frac{1}{x} +\frac{1}{2}\frac{1}{x-1} - \frac{1}{2}\frac{1}{x-2}\\\nonumber \frac{1}{y}y' &=& \frac{(x-1)(x-2) + x(x-2) - x(x-1)}{2x(x-1)(x-2)}\\\nonumber \frac{1}{y}y' &=& \frac{1}{2}\left(\frac{x^2 -2x-x+2+x^2 -2x -x^2 +x}{x(x-1)(x-2)}\right)\\\nonumber \frac{1}{y}y'&=& \frac{1}{2}\left(\frac{x^2 -4x+2}{x(x-1)(x-2)}\right)\\\nonumber y'&=& \sqrt{\frac{x(x-1)}{x-2}} \frac{1}{2}\frac{x^2 -4x +2}{x(x-1)(x-2)}\\\nonumber y'&=& \frac{x^2 -4x+2}{2\sqrt{x(x-1)(x-2)^3}}\end{eqnarray} - Example 6 Calculate the \(y'\) of a function \(y = x \sqrt[3]{\frac{x^2}{x^2 +1}}\)
Solution \begin{eqnarray}y &=& x \sqrt[3]{\frac{x^2}{x^2 +1}}/\ln \\\nonumber \ln y&=& \ln x + \frac{1}{3}\ln x^2 -\frac{1}{3}\ln(x^2 +1) \\\nonumber \frac{1}{y} y' &=& \frac{1}{x} + \frac{2}{3x} - \frac{1}{3}\frac{2x}{x^2 + 1} \\\nonumber \frac{1}{y}y'&=& \frac{5}{3x} - \frac{1}{3}\frac{2x}{x^2 +1}\\\nonumber \frac{1}{y}y'&=& \frac{5}{3x} - \frac{2x}{3(x^2+1)}/\cdot y \\\nonumber y'&=& x \sqrt[3]{\frac{x^2}{x^2 +1}}\left(\frac{5x^2 -5 -2x^2}{3x(x^2 +1)}\right)\\\nonumber y'&=& \frac{5x^2 +5}{3(x^2+1)}\sqrt[3]{\frac{x^2}{x^2 +1}}\end{eqnarray} - Example 7 Calculate the \(y'\) of a function \(y = \frac{(x-2)^9}{\sqrt{(x-1)^5(x-3)^{11}}}\)
Solution \begin{eqnarray}y &=& \frac{(x-2)^9}{\sqrt{(x-1)^5(x-3)^{11}}}/\ln \\\nonumber \ln y &=& 9\ln(x-2) - \frac{5}{2}\ln(x-1) - \frac{11}{2}\ln(x-3)/' \\\nonumber \frac{1}{y}y' &=& \frac{9}{x-2} - \frac{5}{2(x-1)} - \frac{11}{2(x-3)}\\\nonumber \frac{1}{y}y'&=& \frac{18(x-1)(x-3) - 5(x-2)(x-3) - 11(x-1)(x-2)}{2(x-2)(x-3)(x-1)}\\\nonumber \frac{1}{y} y'&=& \frac{2(x^2 - 7x + 1)}{2(x-1)(x-2)(x-3)}/\cdot y\\\nonumber y'&=& \frac{(x-2)^9}{\sqrt{(x-1)^5(x-3)^{11}}} \cdot \frac{x^2 - 7x + 1}{(x-1)(x-2)(x-3)} \\\nonumber y'&=& \frac{(x-2)^8(x^2 - 7x + 1)}{\sqrt{(x-1)^7(x-3)^{13}}}\end{eqnarray} - Example 8 Calculate the \(y'\) of a function \(y = x^x\)
Solution \begin{eqnarray}y &=& x^x/\ln \\\nonumber \ln y &=& x\ln x/' \\\nonumber \frac{1}{y}y'&=& \ln x + x\frac{1}{x} = \ln x + 1/y\\\nonumber y' &=& x^x \ln x + x^x\end{eqnarray} - Example 9 Calculate the \(y'\) of a function \(y = x^{x^2}\)
Solution \begin{eqnarray}y &=& x^{x^2}/\ln \\\nonumber \ln y &=& x^2 \ln x/'\\\nonumber \frac{1}{y}y' &=& 2x\ln x + x^2 \frac{1}{x}\\\nonumber \frac{1}{y}y' &=& 2x\ln x + x /: y \\\nonumber y' &=& x^{x^2 +1} (2\ln x + 1)\end{eqnarray} - Example 10 Calculate the \(y'\) of a function \(y = \sqrt[x]{x}=x^{\frac{1}{x}}\)
Solution \begin{eqnarray}y &=& x^{\frac{1}{x}}/\ln \\\nonumber \ln y &=& \frac{1}{x} \ln x/'\\\nonumber \frac{1}{y}y' &=& -\frac{1}{x^2}\ln x + \frac{1}{x^2}\\\nonumber \frac{1}{y}y'&=& \frac{1}{x^2}(1-\ln x)/\cdot y \\\nonumber y'&=& x^{\frac{1}{x}-2}(1-\ln x)\end{eqnarray}
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