Question

How do you find the derivative of $\dfrac{\text{d}}{\text{dx}}\dfrac{\left( 2x + 1 \right)}{x^{2} – 1}$ ?

Answer 1

In this question, we need to find the derivative of $\dfrac{\text{d}}{\text{dx}}\dfrac{\left( 2x + 1 \right)}{x^{2} – 1}$ . Mathematically, a derivative is defined as a rate of change of function with respect to an independent variable given in the function. Let us consider the given expression as $y$ , the expression $y$ is in the form of $\dfrac{u}{v}$ . Since the expression is in the form of $\dfrac{u}{v}$ , we need to use the quotient rule to differentiate the given expression. First we need to differentiate $u$ and then $v$ . Then we need to substitute the values in the quotient rule to find the derivative of the given expression. With the help of quotient rules and derivative rules, we can easily find the derivative of the given expression.
Quotient rule :
The quotient rule is nothing but a method used in finding the derivative of a function which is the ratio of two differentiable functions.
Let $y = \dfrac{u}{v}$ , then the derivative of $y$ is
$\dfrac{\text{dy}}{\text{dx}} = \dfrac{\left( v\left( \dfrac{\text{du}}{\text{dx}} \right) – u\left( \dfrac{\text{dv}}{\text{dx}} \right) \right)}{v^{2}}$
Where,
$\dfrac{\text{dy}}{\text{dx}}$ is the derivative of $y$ with respect to $x$
$v$ is the variable
$\dfrac{\text{dv}}{\text{dx}}$ is the derivative of $v$ with respect to $x$
$u$ is the variable
$\dfrac{\text{du}}{\text{dx}}$ is the derivative of $u$ with respect to $x$ .
Derivative rules used :
1. $\dfrac{d}{\text{dx}}\left( x^{n} \right) = nx^{n – 1}$
2. $\dfrac{d}{{dx}}\left( k \right) = 0$
3. $\dfrac{\text{d}}{\text{dx}}\left( \text{kx} \right) = x$

Complete answer:
Given, $\dfrac{\text{d}}{\text{dx}}\dfrac{2x + 1}{x^{2} – 1}$
Let us assume that $y = \dfrac{2x + 1}{x^{2} – 1}$ which is in the form of $y = \dfrac{u}{v}$
We can differentiate the given expression with the help of quotient rule.
$\dfrac{\text{dy}}{\text{dx}} = \dfrac{\left( v\left( \dfrac{\text{du}}{\text{dx}} \right) – u\left( \dfrac{\text{dv}}{\text{dx}} \right) \right)}{v^{2}}$ ••• (1)
Let $u = 2x + 1$ and $v = x^{2} – 1$
Now we can differentiate $u$ with respect to $x$ ,
$\dfrac{\text{du}}{\text{dx}} = \dfrac{\text{d}}{\text{dx}}\left( 2x + 1 \right)$
On differentiating,
We get,
$\dfrac{\text{du}}{\text{dx}} = 2$
Then we can differentiate $v$ with respect to $x$ ,
$\dfrac{\text{dv}}{\text{dx}} = \dfrac{\text{d}}{\text{dx}}\left( x^{2} – 1 \right)$
On differentiating,
We get,
$\dfrac{\text{dv}}{\text{dx}} = 2x$
By substituting the values in equation (1) ,
We get
$\dfrac{\text{dy}}{\text{dx}} = \dfrac{\left( \left( x^{2} – 1 \right)\left( 2 \right)\left( 2x + 1 \right)\left( 2x \right) \right)}{\left( x^{2} – 1 \right)^{2}}$
On simplifying,
We get,
$\dfrac{\text{dy}}{\text{dx}} = \dfrac{\left( 2x^{2} – 2 \right) - \left( 4x^{2} + 2x \right)}{\left( x^{2} – 1 \right)^{2}}$
$\Rightarrow \dfrac{\text{dy}}{\text{dx}} = \dfrac{(2x^{2} – 2 – 4x^{2} – 2x)}{\left( x^{2} – 1 \right)^{2}}$
On further simplifying,
We get,
$\dfrac{\text{dy}}{\text{dx}} = \dfrac{\left( - 2x^{2} – 2x – 2 \right)}{\left( x^{2} – 1 \right)^{2}}$
By taking $- 2$ common from the numerator,
We get,
$\dfrac{\text{dy}}{\text{dx}} = - \dfrac{2\left( x^{2} + x + 1 \right)}{\left( x^{2} – 1 \right)^{2}}$
Thus we get the derivative of $\dfrac{\text{d}}{\text{dx}}\dfrac{\left( 2x + 1 \right)}{x^{2} – 1}$ is $- \dfrac{2\left( x^{2} + x + 1 \right)}{\left( x^{2} – 1 \right)^{2}}$ .
The derivative of $\dfrac{\text{d}}{\text{dx}}\dfrac{\left( 2x + 1 \right)}{x^{2} – 1}$ is $- \dfrac{2\left( x^{2} + x + 1 \right)}{\left( x^{2} – 1 \right)^{2}}$.

Note:
Mathematically , Derivative helps in solving the problems in calculus and in differential equations. The derivative of $y$ with respect to $x$ is represented as $\dfrac{\text{dy}}{\text{dx}}$ . Here the notation $\dfrac{\text{dy}}{\text{dx}}$ is known as Leibniz's notation .A simple example for a derivative is the derivative of $x^{3}$ is $3x$ . Derivative is applicable in trigonometric functions also . While opening the brackets make sure that we are opening the brackets properly with their respective signs.Also, while differentiating we should be careful in using the power rule $\dfrac{d}{\text{dx}}\left( x^{n} \right) = nx^{n – 1}$ , a simple error that may happen while calculating.