Question

Differentiate: $\log \left( {x + \dfrac{1}{x}} \right)$
A. $x + \dfrac{1}{x}$
B. $\dfrac{{1 + \dfrac{1}{{{x^2}}}}}{{x + \dfrac{1}{x}}}$
C. $\dfrac{{1 - \dfrac{1}{{{x^2}}}}}{{x + \dfrac{1}{x}}}$
D. $1 + \dfrac{1}{x}$

Answer 1

Hint: The given function is a logarithmic function. Substitute $y = x + \dfrac{1}{x}$. Then the given function is a function of $y$ and $y$ is again a function of $x$. So, find the differentiation of the given function using chain rule.

Formula Used:
If $z$ is a function of $y$ and $y$ is a function of $x$, then the chain rule for differentiation of $z$ with respect to $x$ is given by $\dfrac{{dz}}{{dx}} = \dfrac{{dz}}{{dy}} \cdot \dfrac{{dy}}{{dx}}$
$\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}$
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$

Complete step by step solution:
The given function is $\log \left( {x + \dfrac{1}{x}} \right)$
Let $f\left( x \right) = \log \left( {x + \dfrac{1}{x}} \right)$
Let us substitute $x + \dfrac{1}{x} = y - - - - - \left( i \right)$
Then the given function becomes $f\left( x \right) = g\left( y \right) = \log y - - - - - \left( {ii} \right)$
According to the chain rule, we have
$\dfrac{d}{{dx}}\left\{ {f\left( x \right)} \right\} = \dfrac{d}{{dy}}\left\{ {g\left( y \right)} \right\} \cdot \dfrac{{dy}}{{dx}} - - - - - \left( {iii} \right)$
Differentiating both sides of the equation $\left( i \right)$ with respect to $x$, we get
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {x + \dfrac{1}{x}} \right)$
Differentiation of the sum of some functions is equal to the sum of the differentiation of the functions.
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right)$
Now, $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$
Putting $n = 1$, we get $\dfrac{d}{{dx}}\left( x \right) = 1{x^{1 - 1}} = {x^0} = 1$
Putting $n = - 1$, we get $\dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right) = \dfrac{d}{{dx}}\left( {{x^{ - 1}}} \right) = - 1{x^{ - 1 - 1}} = - {x^{ - 2}} = - \dfrac{1}{{{x^2}}}$
So, $\dfrac{{dy}}{{dx}} = \left( 1 \right) + \left( { - \dfrac{1}{{{x^2}}}} \right) = 1 - \dfrac{1}{{{x^2}}}$
Differentiating the function $g\left( y \right)$ with respect to $y$, we get
$\dfrac{d}{{dy}}\left\{ {g\left( y \right)} \right\} = \dfrac{d}{{dy}}\left\{ {\log y} \right\} = \dfrac{1}{y}$
Now, substituting these expressions in equation $\left( {iii} \right)$, we get
$\dfrac{d}{{dx}}\left\{ {f\left( x \right)} \right\} = \left( {\dfrac{1}{y}} \right)\left( {1 - \dfrac{1}{{{x^2}}}} \right) = \dfrac{{1 - \dfrac{1}{{{x^2}}}}}{y}$
Substituting $y = x + \dfrac{1}{x}$ from equation $\left( i \right)$, we get
$\dfrac{d}{{dx}}\left\{ {f\left( x \right)} \right\} = \dfrac{{\left( {1 - \dfrac{1}{{{x^2}}}} \right)}}{{\left( {x + \dfrac{1}{x}} \right)}}$

Option ‘C’ is correct

Note: Chain rule is used for such types of functions which are compositions of other functions. In this method, we need to substitute for the function for which the given function becomes critical.