Question

How do I find the derivative of a fraction?

Answer 1

For finding the derivative of a fraction, we will use the quotient rule to differentiate the fraction or any other fraction which are written as quotient or fraction of two functions or expressions.

Formula used:
Quotient rule,
If $f\left( x \right) = \dfrac{{g\left( x \right)}}{{h\left( x \right)}}$
Then, $\dfrac{{df}}{{dx}} = \dfrac{{\dfrac{{dg}}{{dx}} \times h\left( x \right) - \dfrac{{dh}}{{dx}} \times g\left( x \right)}}{{{{\left( {h\left( x \right)} \right)}^2}}}$
Here,
$g\left( x \right),h\left( x \right)$ , will be the two functions.
$\dfrac{{dg}}{{dx}}$ , will be the function differentiable at $g$ with respect to $x$
$\dfrac{{dh}}{{dx}}$ , will be the function differentiable at $h$ with respect to $x$

Complete Step by Step Solution:
With an example, we will show how to differentiate the fraction. So let us take a function $f\left( x \right) = \dfrac{{3 - 2x - {x^2}}}{{{x^2} - 1}}$ . Here, $g\left( x \right)$ will be equal to $3 - 2x - {x^2}$ and $h\left( x \right)$ will be equal to ${x^2} - 1$ .
Since, $g\left( x \right) = 3 - 2x - {x^2}$
Therefore, $\dfrac{{dg}}{{dx}} = - 2 - 2x$
Similarly, we have $h\left( x \right) = {x^2} - 1$
$\Rightarrow \dfrac{{dh}}{{dx}} = 2x$
So now substituting these values, in the equation we get
$\dfrac{{df}}{{dx}} = \dfrac{{\left( {2 - 2x} \right) \times \left( {{x^2} - 1} \right) - 2x \times \left( {3 - 2x - {x^2}} \right)}}{{{{\left( {{x^2} - 1} \right)}^2}}}$

Now on solving the braces of the right side of the equation, we get
$\Rightarrow \dfrac{{ - 2{x^3} - 2{x^2} + 2x + x - 6x + 4{x^2} + 2{x^3}}}{{{{\left( {{x^2} - 1} \right)}^2}}}$
And on solving the above equation, we get
$\Rightarrow \dfrac{{2{x^2} - 4x + 2}}{{{{\left( {{x^2} - 1} \right)}^2}}}$
And since, the above equation follows the algebraic formula, so we can write it as
$\Rightarrow \dfrac{{2{{\left( {x - 1} \right)}^2}}}{{{{\left( {{x^2} - 1} \right)}^2}}}$
So by canceling the like terms, we can write it as
$\Rightarrow \dfrac{2}{{{{\left( {{x^2} - 1} \right)}^2}}}$
And hence, in this, we can solve the derivative for the fractions.

Note:
For the quotient rule there will be the requirement of two functions $f$ and $g$ , in which both of them are defined in a neighborhood of some point $a$ and differentiable at $a$ , with $g\left( a \right) \ne 0$ .
Since $g\left( a \right) \ne 0$ and $g$ is continuous at $a$ , then we know that there exists $\delta > 0$ such that $g\left( a \right) \ne 0$ for $\left| {x - a} \right| < \delta$ .
Therefore the function $F\left( x \right) = \dfrac{{f\left( x \right)}}{{g\left( x \right)}}$ is defined in a neighborhood of $a$ and we can ask ourselves if it is differentiable at $a$ and we will compute its derivative. So this is all the idea about the differentiation.