The quotient rule is a fundamental rule in differential calculus. It is one of the basic, simple and widely used rule to differentiate equations. This rule is used to differentiate functions that have two differentiable equations, one in the numerator and the other in the denominator.
Index
History
The quotient rule is a fundamental rule in differentiating functions that are of the form numerator divided by the denominator in calculus.
This rule bears a lot of similarity to another well-known rule in calculus called the product rule.
Gottfried Wilhelm Leibniz was one of the most important German logicians, mathematicians and natural philosophers during the age of enlightenment, during which the whole world was influenced by the ideas of Europeans.
The notations for integration \(\int\) and differentiation \(d\) were defined by him. He also proposed many theories of calculus like the fundamental theorem of calculus, Leibniz integral rule and Leibniz’s law.
Leibniz is credited alongside Newton for the invention of calculus. Leibniz also proposed many important things in fields like probability theory, biology, geology, psychology and computer science.
The Quotient Rule Explained
When differentiating functions which are the form \(\frac{f(x)}{g(x)}\). First differentiate the function in numerator(\(f(x)\)) and multiply with the denominator function, then subtract this with the differentiation of function in the denominator multiplied with the function in the numerator, now divide this whole expression by the square of the denominator function, under the conditions that \(f(x)\) and \(g(x)\) are both differentiable functions and \(g(x) \neq 0\).
In mathematical terms this can be represented as,
If \(h(x) = \frac{f(x)}{g(x)}\)
\(\large{h'(x) = \frac{g(x)f'(x) – f(x)g'(x)}{(g(x))^2}}\)
Proving the Quotient Rule
There are many ways to prove this rule, some of these are using the definition of derivative, using product rule and using chain rule.
Using the Definition of Derivative
\(h(x) = \frac{f(x)}{g(x)}\)
Differentiating with respect to \(x\) on both sides
\(\Rightarrow h'(x) = \lim_{s \rightarrow 0} \frac{h(x + s) – h(x)}{s}\)
\(\Rightarrow h'(x) = \lim_{s \rightarrow 0} \frac{f(x + s)g(x) – g(x + s)f(x)}{sg(x + s)g(x)}\)
Adding and subtracting \(g(x)f(x)\) to the numerator
= \( \lim_{s \rightarrow 0} \frac{f(x + s)g(x) – g(x)f(x) + g(x)f(x) – g(x + s)f(x)}{s} \times \frac{1}{(g(x))^2}\)
= \(\left( g(x) \lim_{s \rightarrow 0} \frac{f(x + s) – f(x)}{s} – f(x) \lim_{s \rightarrow 0} \frac{g(x + s) – g(x)}{s} \right) \times \frac{1}{(g(x))^2}\)
= \(\frac{1}{(g(x))^2} \times (g(x)f'(x) – g'(x)f(x))\)
\(\Rightarrow \large{h'(x) = \frac{g(x)f'(x) – f(x)g'(x)}{(g(x))^2}}\)
Using Product and Chain Rules
\(h(x) = \frac{f(x)}{g(x)}\)
\(\Rightarrow h(x) = f(x)(g(x))^{-1}\)
Differentiating with respect to \(x\) on both sides
\(h'(x) = f'(x)(g(x))^{-1} + f(x)\left[ (g(x))^{-1} \right]’\)
As we know that \(\frac{d(x^n)}{dx} = n.x^{n-1}\)
= \(f'(x)(g(x))^{-1} + f(x) \left[ -(g(x))^{-1-1} \times g'(x) \right]\)
= \(\frac{f'(x)}{g(x)} – \frac{f(x)g'(x)}{(g(x))^2}\)
\(\Rightarrow \large{h'(x) = \frac{g(x)f'(x) – f(x)g'(x)}{(g(x))^2}}\)
Using only Product Rule
\(h(x) = \frac{f(x)}{g(x)}\)
\(f(x) = h(x)g(x)\)
Differentiating on both side with respect to \(x\)
\(f'(x) = h'(x)g(x) + h(x)g'(x)\)
Substituting \(h(x)\)
\(\Rightarrow f'(x) = h'(x)g(x) + \frac{f(x)}{g(x)}g'(x)\)
\(\Rightarrow h'(x)g(x) = f'(x) – \frac{f(x)}{g(x)}g'(x)\)
\(\Rightarrow h'(x)g(x) = \frac{f'(x)g(x) – f(x)g'(x)}{g(x)}\)
\(\Rightarrow \large{h'(x) = \frac{f'(x)g(x) – f(x)g'(x)}{(g(x))^2}}\)
The above are some of the ways to prove this rule.
Applications of Quotient Rule
- Mainly used to calculate the derivative of the functions which are in the forms of fractions.
- This rule can be used as a substitute for the product rule when calculating the derivative of a function in the form of a fraction.
- It can be used to find the rate at which a tank or pool is filled with water.
- It provides a better alternative to the product rule in some functions where applying the product rule directly could be lengthy, but by taking the reciprocal of one of the functions can make differentiating it easy.
Example Problems
Question 1. Differentiate the function \(y = \frac{x^3 + 3x}{x^2 + x}\).
Solution. Given, \(y = \frac{x^3 + 3x}{x^2 + x}\)
\(y’ = \left[ \frac{x^3 + 3x}{x^2 + x} \right]’\)
Using quotient rule, \(h'(x) = \frac{f'(x)g(x) – f(x)g'(x)}{(g(x))^2}\) on the given equation
\(y’ = \frac{(x^3 + 3x)'(x^2 + x) – (x^3 + 3x)(x^2 + x)’}{(x^2 + x)^2}\)
= \(\frac{(3x^2 + 3)(x^2 + x) – (x^3 + 3x)(2x + 1)}{(x^2 + x)^2}\)
= \(\frac{3x^4 + 3x^3 + 3x^2 + 3x – (2x^4 + x^3 + 6x^2 + 3x)}{x^4 + x^2 + 2x^3}\)
\(y’ = \frac{x^4 + 2x^3 – 3x^2}{x^4 + x^2 + 2x^3}\)
Question 2. Differentiate the following equation \(y = \frac{\sin(x^3 + 3x^2)}{\sin(2x + 9)}\).
Solution. Given, \(y = \frac{\sin(x^3 + 3x^2)}{\sin(2x + 9)}\)
\(y’ = \left[ \frac{\sin(x^3 + 3x^2)}{\sin(2x + 9)} \right]’\)
Using quotient rule, \(h'(x) = \frac{f'(x)g(x) – f(x)g'(x)}{(g(x))^2}\) on the given equation
\(y’ = \frac{(\sin(2x + 9)) (\sin(x^3 + 3x^2))’ – (\sin(2x + 9))’ (\sin(x^3 + 3x^2))}{(\sin(2x + 9))^2}\)
= \(\frac{(\sin(2x + 9)) ((3x^2 + 6x) \cos(x^3 + 3x^2)) – (2 \cos(2x + 9)) (\sin(x^3 + 3x^2))}{\sin^2(2x + 9)}\)
= \(\frac{((3x^2 + 6x) \sin(2x + 9) \cos(x^3 + 3x^2)) – (2 \sin(x^3 + 3x^2) \cos(2x + 9))}{\sin^2(2x + 9)}\)
\(y’ = (3x^2 + 6x)\frac{\cos(x^3 + 3x^2)}{\sin(2x + 9)} – 2 \sin(x^3 + 3x^2)\frac{\cot(2x + 9)}{\sin(2x + 9)}\)
FAQs
No, the quotient rule is a fundamental rule of differential calculus. But this rule can be proved using the product.
There is often a misconception that quotient rule and product rule are the same, but this is not true they can only be used interchangeably due to their similarity in the format of functions.
The quotient rule in calculus, is the formula used to differentiate the functions which have two differentiable functions in the numerator and denominator.
No, the quotient law of limits is that, limits can be applied to functions in the numerator and denominator separately, whereas the quotient law of calculus is for differentiating functions with differentiable numerator and denominator functions.
Yes, the quotient rule and product rule can be used interchangeably, but the efficiency and easy of problem solving is largely affected based on the rule one uses.
