The quotient rule

Differentiation: Quotient rule

The quotient rule

We saw earlier that we can multiply and compose functions. We can also divide functions. We call the result the quotient. The quotient of the functions #\blue{g(x)}=\blue{x^3+1}# and #\green{h(x)}=\green{x+1}# is the function #f(x)=\frac{\blue{x^3+1}}{\green{x+1}}#. In this case, same as with regular fractions, we call #\blue{g(x)}# the numerator and #\green{h(x)}# the denominator.

We can calculate the derivative of a quotient by means of the quotient rule.

Quotient rule

For the quotient of two functions \[f(x)=\dfrac{\blue{g(x)}}{\green{h(x)}}\] the following applies:

\[f'(x)=\dfrac{\green{h(x)}\cdot \orange{g'(x)}-\blue{g(x)}\cdot \purple{h'(x)}}{(\green{h(x)})^2}\]

Example

\[f(x)=\dfrac{\blue{2x}}{\green{x+1}}\] gives \[f'(x)=\dfrac{(\green{x+1})\cdot \orange{2}-\blue{2x}\cdot \purple{1}}{(\green{x+1})^2
}\]

We can easily prove the quotient rule with the aid of the product rule. Suppose we have a function #f# of the form #f=\frac{g}{h}#. When we multiply both sides by #h#, we get #g=h\cdot f#. When we apply the product rule to this, we get
\[g'=h'\cdot f + h\cdot f'\]
We can rewrite this to
\[h\cdot f'=g'-h'\cdot f\]
Dividng both sides by #h# gives
\[f'=\frac{g'-h'\cdot f}{h}\]
After which we can substitute #f=\frac{g}{h}#
\[f'=\frac{g'-h'\cdot \dfrac{g}{h}}{h}\]
Now we multiply the numerator and denominator by #h#
\[f'=\frac{g'\cdot h-h'\cdot g}{h^2}\]
This is the quotient rule.

We can follow the step-by-step guide below to apply the quotient rule.

Step-by-step guide quotient rule

	Step-by-step	Example
	Consider the #f(x)#, which is a quotient of two functions.	#\dfrac{\sin(x)}{x^2+1}#
Step 1	Distinguish the numerator #\blue{g(x)}# and the denominator #\green{h(x)}#.	#\blue{g(x)}=\blue{\sin(x)}# #\green{h(x)}=\green{x^2+1}#
Step 2	Calculate #\orange{g'(x)}# and #\purple{h'(x)}#.	#\orange{g'(x)}=\orange{\cos(x)}# #\purple{h'(x)}=\purple{2x}#
Step 3	Calculate the derivative of #f# using the formula: \[f'(x)=\dfrac{\green{h(x)}\cdot \orange{g'(x)}-\blue{g(x)}\cdot \purple{h'(x)}}{(\green{h(x)})^2}\]	#\dfrac{(\green{x^2+1})\cdot \orange{\cos(x)}-\blue{\sin(x)}\cdot \purple{2x}}{(\green{x^2+1})^2}#

Calculate the derivative of #f(x)=\frac{2\cdot x^2+2}{4\cdot x^4+1}#.

#f'(x)=# #{{-16\cdot x^5-32\cdot x^3+4\cdot x}\over{\left(2\cdot x^2-2\cdot x+1\right)^2\cdot \left(2\cdot x^2+2\cdot x+1\right)^2}}#

Step 1	We determine #g(x)# and #h(x)# such that #f(x)=\frac{g(x)}{h(x)}#. #\begin{array}{rcl} g(x)&=&2\cdot x^2+2\\ h(x)&=&4\cdot x^4+1\end{array}#
Step 2	We calculate the derivative #g'(x)# and #h'(x)#. #\begin{array}{rcl} g'(x)&=&\dfrac{\dd}{\dd x}\left(2\cdot x^2+2\right)\\ &&\blue{\text{definition derivative}}\\ &=&4\cdot x\\ &&\blue{\text{sum rule, power rule and constant rule}}\end{array}# #\begin{array}{rcl} h'(x)&=&\dfrac{\dd}{\dd x}\left(4\cdot x^4+1\right)\\ &&\blue{\text{definition derivative}}\\ &=&16\cdot x^3\\ &&\blue{\text{sum rule, power rule and constant rule}}\end{array}#
Step 3	#\begin{array}{rcl} \dfrac{\dd}{\dd x}f(x)&=&\dfrac{h(x)\cdot g'(x)-g(x)\cdot h'(x)}{h(x)^2}\\ &&\blue{\text{quotient rule}}\\ &=&\dfrac{4\cdot x\cdot (4\cdot x^4+1)-16\cdot x^3\cdot (2\cdot x^2+2)}{(4\cdot x^4+1)^2}\\ &&\blue{\text{substituted}}\\ &=&\displaystyle {{-16\cdot x^5-32\cdot x^3+4\cdot x}\over{\left(2\cdot x^2-2\cdot x+1\right)^2\cdot \left(2\cdot x^2+2\cdot x+1\right)^2}}\\ &&\blue{\text{simplified}} \end{array}#

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