Question

How do you find the derivative of $\ln \left( 1+\left( \dfrac{1}{x} \right) \right)$?

Answer 1

This question is from the topic of differentiation of calculus chapter. In this question, we are going to find the derivative. We are going to use formulas of differentiation like $\dfrac{d}{dx}\ln \left( x \right)=\dfrac{1}{x}$ and $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$. We will use chain rule here. The chain rule is used to differentiate the composite functions. The composite functions should be in the form of f(g(x)), where f(x) and g(x) are two different functions.

Complete step by step answer:
Let us solve this question.
In this question, we have to find the derivative $\ln \left( 1+\left( \dfrac{1}{x} \right) \right)$ or we can say we have to find the differentiation of $\ln \left( 1+\left( \dfrac{1}{x} \right) \right)$.
So, the differentiation of $\ln \left( 1+\left( \dfrac{1}{x} \right) \right)$ will be
$\dfrac{d}{dx}\ln \left( 1+\left( \dfrac{1}{x} \right) \right)$
As we know that $\dfrac{d}{dx}\ln \left( x \right)=\dfrac{1}{x}$.
So, we can write
$\Rightarrow \dfrac{d}{dx}\ln \left( 1+\left( \dfrac{1}{x} \right) \right)=\dfrac{1}{1+\left( \dfrac{1}{x} \right)}\dfrac{d}{dx}\left(1+ \dfrac{1}{x} \right)$
We have used chain rule in the above equation. The chain rule states that the derivative of $f\left( g\left( x \right) \right)$
Is $f'\left( g\left( x \right) \right)\times g'\left( x \right)$. The chain rule helps to differentiate composite functions. Here, we can see that $\ln \left( 1+\left( \dfrac{1}{x} \right) \right)$ is a composite function which is in the form of $f\left( g\left( x \right) \right)$ where g(x) is the function of x that is $\dfrac{1}{x}$ here and f(x) is the function of $\ln$ (that is log base e).
Till now, we have got the equation of differentiation as
$\Rightarrow \dfrac{d}{dx}\ln \left( 1+\left( \dfrac{1}{x} \right) \right)=\dfrac{1}{1+\left( \dfrac{1}{x} \right)}\dfrac{d}{dx}\left(1+ \dfrac{1}{x} \right)$
We will use the formula $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ in the above equation. This formula can be written as $\dfrac{d}{dx}\left( \dfrac{1}{x} \right)=\dfrac{d}{dx}\left( {{x}^{-1}} \right)=(-1){{x}^{-1-1}}=-{{x}^{-2}}=-\dfrac{1}{{{x}^{2}}}$
So, we can write the equation of differentiation as
$\Rightarrow \dfrac{d}{dx}\ln \left( 1+\left( \dfrac{1}{x} \right) \right)=\dfrac{1}{1+\left( \dfrac{1}{x} \right)}\left( -\dfrac{1}{{{x}^{2}}} \right)$
The above equation can also be written as
$\Rightarrow \dfrac{d}{dx}\ln \left( 1+\left( \dfrac{1}{x} \right) \right)=\dfrac{x}{x+1}\left( -\dfrac{1}{{{x}^{2}}} \right)=\dfrac{x}{x+1}\left( -\dfrac{1}{x\times x} \right)=\dfrac{1}{x+1}\left( -\dfrac{1}{x} \right)=-\left( \dfrac{1}{x(x+1)} \right)$
The above equation can also be written as
$\Rightarrow \dfrac{d}{dx}\ln \left( 1+\left( \dfrac{1}{x} \right) \right)=\dfrac{-1}{{{x}^{2}}+x}$

Hence, we get that the derivative of the term $\ln \left( 1+\left( \dfrac{1}{x} \right) \right)$ is $\dfrac{-1}{{{x}^{2}}+x}$.

Note: We have an alternate method to solve this question.
We can write $1+\dfrac{1}{x}$ as $\dfrac{x+1}{x}$.
So, $\ln \left( 1+\left( \dfrac{1}{x} \right) \right)$ can be written as $\ln \left( \dfrac{x+1}{x} \right)$
We know that $\ln \left( \dfrac{a}{b} \right)=\ln a-\ln b$
So, $\ln \left( \dfrac{x+1}{x} \right)$ can be written as $\left( \ln \left( x+1 \right)-\ln x \right)$
Now, we will differentiate the term $\left( \ln \left( x+1 \right)-\ln x \right)$
$\dfrac{d}{dx}\left( \ln \left( x+1 \right)-\ln x \right)$
As we know that $\dfrac{d}{dx}\ln \left( x \right)=\dfrac{1}{x}$. So, the differentiation can be written as
$\Rightarrow \dfrac{d}{dx}\left( \ln \left( x+1 \right)-\ln x \right)=\dfrac{1}{x+1}-\dfrac{1}{x}=\dfrac{x-(x+1)}{x(x+1)}$
The above equation can be written as
$\Rightarrow \dfrac{d}{dx}\left( \ln \left( x+1 \right)-\ln x \right)=\dfrac{-1}{x(x+1)}=\dfrac{-1}{{{x}^{2}}+x}$