Question

Differentiate ${\log _x}e$ with respect to $x$

Answer 1

We can use the property of logarithm that, ${\log _a}b = \dfrac{{\ln b}}{{\ln a}}$ to simplify the given expression ${\log _x}e$. We can then differentiate the expression using chain differentiation rules.

Complete step by step solution:
It is known the property of logarithm functions, that ${\log _a}b = \dfrac{{\ln b}}{{\ln a}}$.
Substituting the value $x$ for $a$ and $e$ for $b$ in the equation${\log _a}b = \dfrac{{\ln b}}{{\ln a}}$ to simplify the expression ${\log _x}e$.
${\log _x}e = \dfrac{{\ln e}}{{\ln x}}$
The value of $\ln e$ is equal to 1.
Thus substituting the value 1 for $\ln e$ in the equation ${\log _x}e = \dfrac{{\ln e}}{{\ln x}}$.
${\log _x}e = \dfrac{1}{{\ln x}}$
Differentiating the expression $\dfrac{1}{{\ln x}}$ with respect to $x$.
$\dfrac{{d\left( {{{\left( {\ln x} \right)}^{ - 1}}} \right)}}{{dx}}$
The chain rule of differentiation states that $\dfrac{{d\left( {f\left( {g\left( x \right)} \right)} \right)}}{{dx}} = f'\left( x \right)\dfrac{{d\left( {g\left( x \right)} \right)}}{{dx}}$.
The differentiation formula for the expression ${x^n}$is $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$.
Similarly the differentiation of $\log x$ with respect to $x$ is $\dfrac{1}{x}$.
Simplifying the above expression, we get
$\dfrac{{d\left( {{{\left( {\ln x} \right)}^{ - 1}}} \right)}}{{dx}} = - 1{\left( {\ln x} \right)^{ - 2}}\left( {\dfrac{{d\left( {\ln x} \right)}}{{dx}}} \right)$
$= - 1{\left( {\ln x} \right)^{ - 2}}\dfrac{1}{x}$
The expression $- 1{\left( {\ln x} \right)^{ - 2}}\dfrac{1}{x}$ can be rewritten as
$= - \dfrac{1}{{x{{\ln }^2}x}}$
Thus the differentiation of ${\log _x}e$ with respect to $x$ is $- \dfrac{1}{{x{{\ln }^2}x}}$.

Note: Students can alternatively use quotient rule of differentiation to calculate the derivate of ${\log _x}e = \dfrac{1}{{\ln x}}$ where quotient rules states that, $\dfrac{d}{{dx}}\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \dfrac{{g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}$
Then,
$\dfrac{d}{{dx}}\left( {{{\log }_x}e} \right) = \dfrac{d}{{dx}}\left( {\dfrac{1}{{\ln x}}} \right) \\\ \Rightarrow \dfrac{{\ln x\dfrac{d}{{dx}}\left( 1 \right) - 1\dfrac{d}{{dx}}\left( {\ln x} \right)}}{{{{\left( {\ln x} \right)}^2}}} \\\$
After applying the formula of differentiation, we will get,
$\dfrac{{ - 1\left( {\dfrac{1}{x}} \right)}}{{{{\ln }^2}x}} \\\ \Rightarrow - \dfrac{1}{{x{{\ln }^2}x}} \\\$