Question

How do I find the derivative of a fraction?

Answer 1

Hint:
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 $f\left(x\right)={\displaystyle \frac{g\left(x\right)}{h\left(x\right)}}$
Then, ${\displaystyle \frac{df}{dx}}={\displaystyle \frac{{\displaystyle \frac{dg}{dx}}\times h\left(x\right)-{\displaystyle \frac{dh}{dx}}\times g\left(x\right)}{{\left(h\left(x\right)\right)}^2}}$
Here,
$g\left(x\right),h\left(x\right)$ , will be the two functions.
${\displaystyle \frac{dg}{dx}}$ , will be the function differentiable at $g$ with respect to $x$
${\displaystyle \frac{dh}{dx}}$ , 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 $f\left(x\right)={\displaystyle \frac{3-2x-x^2}{x^2-1}}$ . 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, ${\displaystyle \frac{dg}{dx}}=-2-2x$
Similarly, we have $h\left(x\right)=x^2-1$
$\Rightarrow {\displaystyle \frac{dh}{dx}}=2x$
So now substituting these values, in the equation we get
${\displaystyle \frac{df}{dx}}={\displaystyle \frac{\left(2-2x\right)\times \left(x^2-1\right)-2x\times \left(3-2x-x^2\right)}{{\left(x^2-1\right)}^2}}$

Now on solving the braces of the right side of the equation, we get
$\Rightarrow {\displaystyle \frac{-2x^3-2x^2+2x+x-6x+4x^2+2x^3}{{\left(x^2-1\right)}^2}}$
And on solving the above equation, we get
$$\Rightarrow {\displaystyle \frac{2x^2-4x+2}{{\left(x^2-1\right)}^2}}$$
And since, the above equation follows the algebraic formula, so we can write it as
$$\Rightarrow {\displaystyle \frac{2{\left(x-1\right)}^2}{{\left(x^2-1\right)}^2}}$$
So by canceling the like terms, we can write it as
$\Rightarrow {\displaystyle \frac{2}{{\left(x^2-1\right)}^2}}$
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 $F\left(x\right)={\displaystyle \frac{f\left(x\right)}{g\left(x\right)}}$ 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.