Question

How do you find the derivatives of $x = \ln \left( {xy} \right)$ ?

Answer 1

Hint : Here the given function can be differentiate by two different method, By taking the exponential function on both sides of the equation and by simply using logarithmic property. Otherwise differentiate a function using the properties of logarithms and its inverse to convert the given equation as a function of y and then use the quotient rule of differentiation.

Complete step-by-step answer :
Differentiation is the algebraic method of finding the derivative for a function at any point. The derivative is a concept that is at the root of calculus. ... Either way, both the slope and the instantaneous rate of change are equivalent, and the function to find both of these at any point is called the derivative. It can be represented as $\dfrac{{dy}}{{dx}}$ , where y is the function of x.
Method : 1
Consider the given function
$\Rightarrow \,\,\,\,x = \ln \left( {xy} \right)$
Take exponential on both sides of the equation, then
$\Rightarrow \,\,\,{e^x} = {e^{\ln (xy)}}$
As we know, one of the logarithmic property ${e^{{{\log }_e}x}} = x,\,x > 0$ where natural logarithm $\ln = {\log _e}$ , then
$\Rightarrow \,\,\,\,{e^x} = xy$
$\Rightarrow \,\,\,\,y = \dfrac{{{e^x}}}{x}$
Differentiate with respect to x using quotient rule i.e., $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v.\,du - u.\,dv}}{{{v^2}}}$
$\Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{{x.\,\dfrac{d}{{dx}}\left( {{e^x}} \right) - {e^x}.\dfrac{{dx}}{{dx}}}}{{{x^2}}}$
Using Standard differentiation formula $\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$ and $\dfrac{{dx}}{{dx}} = 1$
$\Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{{x.\,{e^x} - {e^x}.1}}{{{x^2}}}$
Take ${e^x}$ common in numerator
$\Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = {e^x}\left( {\dfrac{{x\, - 1}}{{{x^2}}}} \right)$

Method : 2
Consider the given function
$\Rightarrow \,\,\,\,x = \ln \left( {xy} \right)$
Using the Product rule of logarithms i.e., $\ln (ab) = \ln a + \ln b$ .
$\Rightarrow \,\,\,\,x = \ln \left( x \right) + \ln \left( y \right)$
$\Rightarrow \,\,\,\,\ln \left( y \right) = x - \ln \left( x \right)$
$\Rightarrow \,\,\,\,\ln \left( y \right) = x + \ln 1 - \ln \left( x \right)$ [ $\because \,\,\ln 1 = 0$ ]
Using the property of logarithms $\ln \left( {\dfrac{a}{b}} \right) = \ln a - \ln b$
$\Rightarrow \,\,\,\,\ln \left( y \right) = x + \ln \left( {\dfrac{1}{x}} \right)$
Take exponential on both sides of the equation, then
$\Rightarrow \,\,\,\,{e^{\ln \left( y \right)}} = {e^{x + \ln \left( {\dfrac{1}{x}} \right)}}$
$\Rightarrow \,\,\,\,{e^{\ln \left( y \right)}} = {e^{x\,}}.{e^{\ln \left( {\dfrac{1}{x}} \right)}}$
Using the logarithmic property ${e^{{{\log }_e}x}} = x,\,x > 0$ where natural logarithm $\ln = {\log _e}$ , then
$\Rightarrow \,\,\,\,y = {e^{x\,}}\dfrac{1}{x}$
$\Rightarrow \,\,\,\,y = \dfrac{{{e^{x\,}}}}{x}$
Differentiate with respect to x using quotient rule i.e., $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v.\,du - u.\,dv}}{{{v^2}}}$
$\Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{{x.\,\dfrac{d}{{dx}}\left( {{e^x}} \right) - {e^x}.\dfrac{{dx}}{{dx}}}}{{{x^2}}}$
Using Standard differentiation formula $\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$ and $\dfrac{{dx}}{{dx}} = 1$
$\Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{{x.\,{e^x} - {e^x}.1}}{{{x^2}}}$
Take ${e^x}$ common in numerator
$\Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = {e^x}\left( {\dfrac{{x\, - 1}}{{{x^2}}}} \right)$
Hence, the derivatives of $x = \ln \left( {xy} \right)$ is $\dfrac{{dy}}{{dx}} = {e^x}\left( {\dfrac{{x\, - 1}}{{{x^2}}}} \right)$
So, the correct answer is “ $\dfrac{{dy}}{{dx}} = {e^x}\left( {\dfrac{{x\, - 1}}{{{x^2}}}} \right)$ ”.

Note : The differentiation is defined as the derivative of a function with respect to the independent variable. Here the dependent variable is y and the independent variable is x. The ln means natural logarithm, the properties will be the same for natural and common logarithms.