Quotient rule is one of the subtopics of differentiation in calculus. It is the most important topic of differentiation (a function that is broken down into small functions). It is basically used in a differentiation problem where one function is divided by the other Quotient Rule:
In calculus, quotient rule is defined as the method of finding the derivatives of two functions that are in ratio and are differentiable.
It is expressed as :
The denominator function times
The derivative of the numerator function
The numerator function times
The derivative of the denominator function
The square of the denominator function
In simple words, it is the differentiation of the division function.
It is quite similar to the product rule in calculus. The only difference between the quotient rule and the product rule is in product rule we have the function of type f(x)*g(x) and in quotient rule, we have the function of type f(x)/g(x).
This rule always starts with the denominator function and ends up with the denominator function.
Examples of these functions are six/ex, 3x/x4, e3x/2x, etc.
Example :
Consider a function f(x) which given by,
f(x) = g(x) / t(x)
Then the derivative of the function f(x) is,
f'(x)= [ g(x) / t(x)]’
The quotient rule gives us the derivatives of a function in a unique way. It is the combination of the original function and its derivatives.
If we have a function f(x) in the form of two functions which are differentiable then the function is expressed as,
If f(x) and g(x) are differentiable, then
\[\frac{{d}}{dx}\] \[\frac{{f(x) }}{g(x)}\] = \[\frac{{g(x)f(x)-f(x){g}'(x)}}{[g(x)]^{2}}\] Let u = f(x) and v = g(x) \[\frac{{d}}{dx}\] [ \[\frac{{u}}{v}\] ] = \[\frac{{v\frac{du}{dx} - u \overline{dx}}}{[v]^{2}}\] DV |
The quotient rule formula seems to be complicated but if you remember it and go step by step then it is easier and it saves our time.
We can find the derivatives of higher-order function also with the help of quotient rule in the form of division.
Consider the given function, it should be in the form of division.
Differentiate both sides of the function with respect to something.
Suppose the function on LHS is y equal to some other function of x.
Then derivatives of LHS will be dy/dx
The derivatives on the RHS will be the lower function times the derivative of the upper function minus the upper function times the derivative of the lower function divided by the square of the lower function.
The answer you will get at the end after simplification will be the derivative of the function y which is given to you.
Here we will work out some examples is that you are able to understand clearly with the help of them.
Example 1: Find the derivatives of the function using quotient rule y= \[\frac{{x+2 }}{x^{2}+1}\]
Solution: To find the derivatives we will follow the above steps.
So, we will differentiate both sides with respect to x
\[\bar{dx}\] = \[\frac{{(x^{2}+1) (1) - (x+2) (2x)}}{[x^{3}+1]^{2}}\] by
\[\frac{{x^{2} + 1 -2x^{2}-4x }}{[x^{3}+1]^{2}}\]
= \[\frac{{1-4x-x^{2} }}{(x^{2}+1)^{2}}\]
Example 2: Find the derivative of the function using quotients rule y = \[\frac{{5x+7 }}{2x-13}\]
Solution: We will differentiate both the sides respect to x
\[\bar{dx}\] = \[\frac{{(2x-13) (5)-(5x+7) (2) }}{[2x -13]^{2}}\] by
= \[\frac{{10x-65-10x-14 }}{[2x-13]^{2}}\]
=\[\frac{{-79 }}{[2x-13]^{2}}\]
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