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.