Question

How do you use the quotient rule to differentiate \[\dfrac{\text{d}}{\text{dx}}\dfrac{\left( 2x + 1 \right)}{x^{2} – 1}\] ?

Answer 1

Hint: In this question, we need to find the differentiation of \[\dfrac{\text{d}}{\text{dx}}\dfrac{\left( 2x + 1 \right)}{x^{2} – 1}\] using quotient rule . Mathematically, a differentiation 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 \[\dfrac{u}{v}\] . First we need to differentiate \[u\] and then \[v\] . Then we need to substitute the values in the quotient rule to find the differentiation of the given expression. With the help of quotient rules and derivative rules, we can easily find the differentiation 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
\[\dfrac{\text{dy}}{\text{dx}} = \dfrac{\left( v\left( \dfrac{\text{du}}{\text{dx}} \right) – u\left( \dfrac{\text{dv}}{\text{dx}} \right) \right)}{v^{2}}\]

Complete step-by-step 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 differentiation of \[\dfrac{\text{d}}{\text{dx}}\dfrac{\left( 2x + 1 \right)}{x^{2} – 1}\] is \[- \dfrac{2\left( x^{2} + x + 1 \right)}{\left( x^{2} – 1 \right)^{2}}\] .
Final answer :
The differentiation 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 , Differentiation 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 differentiation is the differentiation of \[x^{3}\] is \[3x\] . Differentiation 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.