The Product Rule is a formula that we can use to differentiate the product of 2 (or more) functions. The Quotient Rule is for the quotient of two functions (one function divided by another). The rules are quite easy to apply. The challenge is often simplifying your answer so that you can find the coordinates of any stationary points. To do this you need to do some careful factorising. It is also important to be confident in using other differentiation techniques, like the Chain Rule.
On this page, you should learn about
the product rule for differentiation the quotient rule for differentiation Product Rule \(y=uv\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\) \(f(x)=g(x)h(x)\\ f'(x)=g(x)h'(x)+h(x)g'(x)\) Quotient Rule \(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \) \(f(x)=\frac { g(x) }{ h(x) } \\ f'(x)=\frac { h(x)\cdot g'(x)-g(x)\cdot h'(x) }{ { \left[ h(x) \right] }^{ 2 } } \)
The following videos will help you understand all the concepts from this page
In the following example we look at using the product rule to find the derivative of a product of 2 simple functions
Given that \(y=x^2lnx\) , find \(\frac{dy}{dx}\)
Applying the product rule is not difficult, but what can be challenging is simplifying the result so that stationary points can be found. In order to do this careful factorising is required. Here is an example that shows how to do it.
\(f(x)=x^3(x-3)^2\)
The function f has three stationary points. Find the x coordinates of these points
The Quotient Rule is fairly straight forward to apply. However, functions that require this rule often require good knowledge of other differentiation techniques, like the Chain Rule. The example below requires some algebraic simplifying techniques also.
Let \(f(x)=\frac{(3x-2)^2}{x^3} \quad,x\neq 0\)
Find \(f'(x)\)
One of the challenging things about the Quotient Rule is that we are often required to simplify our answers and leave in them in a factorised form so that we can find the coordinates of any stationary points. Here is a typical example where the simplifying requires careful algebraic manipulation.
The graph of \(y=\frac{2x}{\sqrt{x-1}} \quad ,x>1\) has a local minimum.
Find the coordinates of this point.
Here is a quiz that practises The Product Rule
Complete the derivatives below with some of the following functions
1 2x² lnx x x² 2x
\(\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\)
Complete the derivatives below with some of the following functions
lnx (x - 3) (x - 3)² 2(x - 3) x² 3x²
\(\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\)
\(\frac{d}{dx}[e^{2x}lnx]=\)
\(\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\)
\(\frac{d}{dx}(ln2x)=\frac{2}{2x}=\frac{1}{x}\)
Consider two functions f g h f g
The table shows valuse for f g h f' g' h'
Example, f g '
Complete the table
h f g
h ' f g + f g '
Here is a quiz that practises The Quotient Rule
Complete the following derivative
\(\frac{d}{dx}(\frac {x}{x+3})=\frac { a\cdot 1-x\cdot b }{ { (x+3) }^{ c } } \)
\(\frac{d}{dx}(\frac {u}{v})=\frac { v\cdot \frac{du}{dx}-u\cdot \frac{dv}{dx} }{ v^2 } \\ \frac{d}{dx}(\frac {x}{x+3})=\frac { (x+3)\cdot 1-x\cdot 1 }{ { (x+3) }^{ c2} } \)
Complete the following derivative
\(\frac{d}{dx}(\frac {e^x}{x})=\frac { a\cdot e^x-e^x }{ b } \)
\(\frac{d}{dx}(\frac {u}{v})=\frac { v\cdot \frac{du}{dx}-u\cdot \frac{dv}{dx} }{ v^2 } \\
\frac{d}{dx}(\frac {e^x}{x})=\frac { x\cdot e^x-e^x }{ x² } \)
Complete the following derivative
\(\frac{d}{dx}(\frac {lnx}{x})=\frac { a - 1 \cdot b }{ c } \)
\(\frac{d}{dx}(\frac {u}{v})=\frac { v\cdot \frac{du}{dx}-u\cdot \frac{dv}{dx} }{ v^2 } \\
\frac{d}{dx}(\frac {lnx}{x})=\frac { x \cdot \frac{1}{x} - 1 \cdot lnx }{ x^2 } \)
Given that \(f(x) =\frac{x^2-1}{x^2+1}\) , then \(f'(x) =\frac{a}{(x^2+1)^2}\)
Find a
\(f'(x) =\frac{2x(x^2+1)-2x(x^2-1)}{(x^2+1)^2}\\
f'(x) =\frac{2x^3+2x-2x^3+2x}{(x^2+1)^2}\\
f'(x) =\frac{4x}{(x^2+1)^2}\)
Find the value of a
\(\frac{d}{dx}(\frac {x}{x-1})=\frac { a }{ (x-1)^2} \)
\(\frac{d}{dx}(\frac {x}{x-1})=\frac { (x-1) \cdot 1-x \cdot1 }{ (x-1)^2} \\
\quad \quad \quad \quad=\frac { x-1-x }{ (x-1)^2}\\
\quad \quad \quad \quad=\frac { -1 }{ (x-1)^2}\)
Find the value of a
\(\frac{d}{dx}(\frac {2e^x-1}{2e^x+1})=\frac { ae^x }{ (2e^x+1)^2} \)
\(\frac{d}{dx}(\frac {2e^x-1}{2e^x+1})=\frac { (2e^x+1)2e^x-2e^x(2e^x-1) }{ (2e^x+1)^2} \\
\quad \quad \quad \quad=\frac { 4e^{2x}+2e^x-4e^{2x}+2e^x }{ (2e^x+1)^2} \\
\quad \quad \quad \quad=\frac { 4e^x }{ (2e^x+1)^2} \\\)
Consider two functions f g \(h(x)=\frac{f(x)}{g(x)}\)
The table shows valuse for f g h f' g' h'
Example, f g '
Complete the table
\(h(1)=\frac{f(1)}{g(1)}=\frac{3}{-1}=-3\)
\(h'(1)=\frac{g(1)f'(1)-g'(1)f(1)}{[g(1)]^2}=\frac{-1(-2)-1\cdot3}{[-1]^2}=-1\)
Let \(y=xe^x\)
a) Find \(\frac{dy}{dx}\)
b) Show that \(\frac{d^2y}{dx^2}=e^x(2+x)\)
c) Find the coordinates of the stationary point and show that it is a local minimum.
a) Use the Product Rule to find \(\frac{dy}{dx}\)
c) Solve \(\frac{dy}{dx}=0\) to find position of stationary point. Show that \(\frac{d^2y}{dx^2}>0\)
Let \(f(x)={ x }^{ 2 }{ (2x-3) }^{ 3 }\)
a) Find \(f'(x)\)
b) The graph of y = f(x) has stationary points at x = 0, x= \(\frac{3}{2}\) and x = a . Find the value of a
a) You need to use the chain rule to differentiate \({ (2x-3) }^{ 3 }\)
b) Factorise your answer to part a)
Let \(f(x)=\frac{lnx}{x},x>0\)
a) Show that \(f'(x)=\frac{1-lnx}{x^2}\)
b) Find \(f''(x)\)
c) The graph of f has a point of inflexion at A. Find the x-coordinate of A.
a) Use the Quotient Rule to find \(f'(x)\)
c) Solve \(f''(x)=0\) to find point of inflexion
Let \(f(x)=tanx\) . A\((\frac{\pi}{3},\sqrt{3})\) is a point that lies on the graph of \(y=f(x)\)
a) Given that \(tanx=\frac{sinx}{cosx}\) find \(f'(x)\)
b) Show that \(f'(\frac{\pi}{3})=4\)
c) Find the equation of the normal to the curve y = f(x) at the point A
d) Show that the normal crosses the y axis at \(\sqrt{3}+\frac{\pi}{12}\)
Hint a) Use the quotient rule to differentiate tanx
c) \(gradient \ of \ normal =-\frac{1}{gradient \ of \ tangent}\)
Let \(f(x)=e^{2x}cosx\)
a) Find \(f'(x)\)
b) Show that \(f''(x)=4f'(x)-5f(x)\)
a) Use the Product Rule to find \(f'(x)\)
b) Use Product Rule again to find \(f''(x)\)
MY PROGRESS
Self-assessment How much of Product and Quotient Rule have you understood?
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