Question

What is the derivative of $\ln \left( 1+\left( \dfrac{1}{x} \right) \right)$ ?

Answer 1

Here, we have $\ln \left( 1+\left( \dfrac{1}{x} \right) \right)$ which cannot be solved through single-step differentiation. Here, we will apply chain rule to first differentiate the outer function, and then proceed on to differentiate the expression contained in the brackets.

Complete step by step answer:
Derivative of a function means to differentiate the function with respect to the variable contained in that function.
If $y=f\left( x \right)$
Derivative of the function will be given by differentiating $y$ with respect to $x$ .i.e., keeping $x$ constant.
$\dfrac{dy}{dx}=\dfrac{d\left( f\left( x \right) \right)}{dx}$
Certain rules are there to remember while differentiating a function,
$\dfrac{d\left( \ln x \right)}{dx}=\dfrac{1}{x}$
$\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$
If we define two differentiable functions $f\left( x \right)$ and $g\left( x \right)$ and assume a function $h\left( x \right)$ such that
$h\left( x \right)=f\circ g\left( x \right)$
Then derivative of such a composite function is given by chain rule
$\begin{aligned} & \dfrac{d\left( h\left( x \right) \right)}{dx}=\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{dx} \\\ & \Rightarrow \dfrac{d\left( h\left( x \right) \right)}{dx}={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right) \\\ \end{aligned}$
Let us consider
$f\left( x \right)=\ln x$ and
$g\left( x \right)=1+\dfrac{1}{x}$
According to the question, we have
$F\left( x \right)=f\circ g\left( x \right)$
Differentiating both sides with respect to $x$ ,
${F}'\left( x \right)={{\left( f\circ g \right)}^{\prime }}\left( x \right)$
Using chain rule as mentioned above
$\begin{aligned} & \dfrac{d\left( F\left( x \right) \right)}{dx}=\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{dx} \\\ & \Rightarrow \dfrac{d\left( F\left( x \right) \right)}{dx}={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right)......\left( 1 \right) \\\ \end{aligned}$
We have $f\left( x \right)=\ln x$
Differentiating both sides with respect to $x$,
${f}'\left( x \right)=\dfrac{1}{x}......\left( 2 \right)$
Similarly, we have
$g\left( x \right)=1+\dfrac{1}{x}$
Differentiating both sides with respect to $x$ ,
${g}'\left( x \right)=\dfrac{d\left( 1 \right)}{dx}+\dfrac{d{{\left( x \right)}^{-1}}}{dx}$
Derivative of a constant is always zero and using
$\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$
Here, $n=-1$
$\begin{aligned} & \Rightarrow {g}'\left( x \right)=\left( -1 \right){{x}^{-1-1}} \\\ & \Rightarrow {g}'\left( x \right)=\dfrac{-1}{{{x}^{2}}}......\left( 3 \right) \\\ \end{aligned}$
Using the results (2) and (3) in equation (1), we get
$\dfrac{d\left( F\left( x \right) \right)}{dx}=\dfrac{1}{g\left( x \right)}\times \left( \dfrac{-1}{{{x}^{2}}} \right)$
$\begin{aligned} & \Rightarrow \dfrac{d\left( F\left( x \right) \right)}{dx}=\dfrac{1}{1+\dfrac{1}{x}}\times \left( \dfrac{-1}{{{x}^{2}}} \right) \\\ & \Rightarrow \dfrac{d\left( F\left( x \right) \right)}{dx}=\dfrac{x}{x+1}\times \left( \dfrac{-1}{{{x}^{2}}} \right) \\\ & \Rightarrow \dfrac{d\left( F\left( x \right) \right)}{dx}=\dfrac{-1}{x\left( x+1 \right)} \\\ \end{aligned}$

Hence, the derivative of $\ln \left( 1+\left( \dfrac{1}{x} \right) \right)$ is $\dfrac{-1}{x\left( x+1 \right)}$

Note: While using chain rule to differentiate a composite function, always proceed the differentiation in a particular manner: start differentiating the outer expression and then move on to differentiate the inner expressions. In this way, no expression would be left out to differentiate and chances of errors will reduce. Also, keep in mind the constant coefficients of the given variable as they come out of the expression as it is.
For example,
\dfrac{d\left( f\left( ax+b \right) \right)}{dx}=a\left\\{ {f}'\left( ax+b \right) \right\\}