Question

How do you find the differential dy of the function $y = \dfrac{{x + 1}}{{2x - 1}}$ ?

Answer 1

Hint : In the given question, we are required to find the differential for the rational function given to us in variable x as $y = \dfrac{{x + 1}}{{2x - 1}}$ . We need to have basic knowledge about differentiation and important rules related to derivatives such as product rule, quotient rule, etc. so as to solve such questions with ease.

Complete step by step solution:
We have to find the differential dy of the given function $y = \dfrac{{x + 1}}{{2x - 1}}$ . Since the given function is a rational function in variable x, we should have knowledge about the quotient rule of differentiation so as to find the derivative of the function given to us and hence find the differential dy of the function.
According to quotient rule of differentiation, $\dfrac{d}{{dx}}\left[ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \dfrac{{g\left( x \right)\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] - f\left( x \right)\dfrac{d}{{dx}}\left[ {g\left( x \right)} \right]}}{{{{\left[ {g\left( x \right)} \right]}^2}}}$
So, $y = \dfrac{{x + 1}}{{2x - 1}}$
Differentiating both sides of the equation with respect to x, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{x + 1}}{{2x - 1}}} \right)$
Using the quotient rule of differentiation, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \left[ {\dfrac{{\left( {2x - 1} \right)\left( 1 \right) - \left( {x + 1} \right)\left( 2 \right)}}{{{{\left( {2x - 1} \right)}^2}}}} \right]$
Opening brackets and simplifying further, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \left[ {\dfrac{{\left( {2x - 1} \right) - \left( {2x + 2} \right)}}{{{{\left( {2x - 1} \right)}^2}}}} \right]$
$\Rightarrow \dfrac{{dy}}{{dx}} = \left[ {\dfrac{{2x - 1 - 2x - 2}}{{{{\left( {2x - 1} \right)}^2}}}} \right]$
Doing calculations and simplifying, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \left[ {\dfrac{{ - 1 - 2}}{{{{\left( {2x - 1} \right)}^2}}}} \right]$
$\Rightarrow \dfrac{{dy}}{{dx}} = \left[ {\dfrac{{ - 3}}{{{{\left( {2x - 1} \right)}^2}}}} \right]$
Shifting $dx$ to the right side of the equation, we get,
$\Rightarrow dy = \left[ {\dfrac{{ - 3}}{{{{\left( {2x - 1} \right)}^2}}}} \right]dx$
So, the value of differential dy for the given function in x, $y = \dfrac{{x + 1}}{{2x - 1}}$ is $dy = \left[ {\dfrac{{ - 3}}{{{{\left( {2x - 1} \right)}^2}}}} \right]dx$ .
So, the correct answer is “ $dy = \left[ {\dfrac{{ - 3}}{{{{\left( {2x - 1} \right)}^2}}}} \right]dx$ ”.

Note : We have to take care while doing calculative steps as there are high chances of mistakes. Also, take care while applying the rules of differentiation to find differentials as they can be a bit confusing. We use the following formulae to solve the given problem
$\dfrac{d}{{dx}}\left[ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \dfrac{{g\left( x \right)\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] - f\left( x \right)\dfrac{d}{{dx}}\left[ {g\left( x \right)} \right]}}{{{{\left[ {g\left( x \right)} \right]}^2}}}$