Question

What is the derivative of \[\ln \left( {2x + 1} \right)\] ?

Answer 1

Hint: Here, the given question has a logarithmic function. We have to find the derivative or differentiated term of the function. First consider the function \[y\] , then differentiate \[y\] with respect to \[x\] by using a standard differentiation formula of the logarithm function and use chain rule for differentiation. And on further simplification we get the required differentiate value.

Complete step-by-step answer:
The differentiation of a function is defined as the derivative or rate of change of a function. The function is said to differentiable if limit exists.
The Chain Rule is a formula for computing the derivative of the composition of two or more functions.
The chain rule expressed as \[\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]
Consider the given function and call it as \[y\]
  \[ \Rightarrow y = \ln \left( {2x + 1} \right)\] ---------- (1)
Differentiate function y with respect to x
 \[ \Rightarrow \dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {\ln \left( {2x + 1} \right)} \right)\] -------(2)
Here, we have to use the chain rule method i.e., \[\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\] to differentiate the above function.
Given function \[y = \ln \left( {2x + 1} \right)\] contains a function i.e., \[2x + 1\] within \[\ln \left( u \right)\] . Letting \[u = 2x + 1\] , now we can apply a chain rule.
Equation (2) can be written as
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{du}}\left( {\ln \left( u \right)} \right) \cdot \dfrac{d}{{dx}}\left( {2x + 1} \right)\] --------(3)
Now, consider
 \[ \Rightarrow \dfrac{d}{{du}}\left( {\ln \left( u \right)} \right)\]
On differentiating using a formula \[\dfrac{d}{{dx}}\ln \left( x \right) = \dfrac{1}{x}\] , we have
 \[ \Rightarrow \dfrac{d}{{du}}\left( {\ln \left( u \right)} \right) = \dfrac{1}{u}\]
Where, \[u = 2x + 1\] on substituting, we get
 \[ \Rightarrow \dfrac{d}{{du}}\left( {\ln \left( u \right)} \right) = \dfrac{1}{{2x + 1}}\] --------(a)

Next, consider
 \[ \Rightarrow \dfrac{d}{{dx}}\left( {2x + 1} \right)\]
 \[ \Rightarrow \dfrac{d}{{dx}}\left( {2x + 1} \right) = \dfrac{d}{{dx}}\left( {2x} \right) + \dfrac{d}{{dx}}\left( 1 \right)\]
On differentiating using a formula \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\] and remember differentiation of constant term is zero, we have
 \[ \Rightarrow \dfrac{d}{{dx}}\left( {2x + 1} \right) = 2 + 0\]
 \[ \Rightarrow \dfrac{d}{{dx}}\left( {2x + 1} \right) = 2\] --------(b)
Substitute (a) and (b) in Equation (3), then
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{2x + 1}} \cdot 2\]
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{2}{{2x + 1}}\]
Hence, it’s a required differentiated value.
So, the correct answer is “ \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{2}{{2x + 1}}\] ”.

Note: Here in this question, we used some standard differentiation formula i.e.,
 \[\dfrac{d}{{dx}}\ln \left( x \right) = \dfrac{1}{x}\]
 \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
 \[\dfrac{d}{{dx}}\left( {constant} \right) = 0\]
The student must know about the differentiation formulas for the logarithm function and these differentiation formulas are standard. If the function is a product of two terms and the both terms are the function of x then we use the product rule of differentiation to the function.