Revise
MME Premium

Supercharge your learning

Learn More

Quotient Rule

A LevelAQAEdexcelOCR

Quotient Rule

You could use the Product Rule here, but that might get a little messy and a bit laborious. Here’s another rule which saves us a lot of time and effort.

A LevelAQAEdexcelOCR
https://www.google.com

Quotient Rule Formula

For a function y=f(x)=u(x)v(x)textcolor{limegreen}{y} = f(textcolor{blue}{x}) = dfrac{u(textcolor{blue}{x})}{v(textcolor{blue}{x})}, we have the derivative (with respect to xtextcolor{blue}{x}) given by

dydx=df(x)dx=v(x)u′(x)−u(x)v′(x)(v(x))2dfrac{dtextcolor{limegreen}{y}}{dtextcolor{blue}{x}} = dfrac{df(textcolor{blue}{x})}{dtextcolor{blue}{x}} = dfrac{v(textcolor{blue}{x})u'(textcolor{blue}{x}) – u(textcolor{blue}{x})v'(textcolor{blue}{x})}{(v(textcolor{blue}{x}))^2}

A LevelAQAEdexcelOCR
A LevelAQAEdexcelOCR
https://www.google.com

Example 1: Using the Quotient Rule

Say we have a function y=exsin⁡xtextcolor{limegreen}{y} = dfrac{e^textcolor{blue}{x}}{sin textcolor{blue}{x}}. Find dydxdfrac{dtextcolor{limegreen}{y}}{dtextcolor{blue}{x}}.

[3 marks]

Let u(x)=exu(textcolor{blue}{x}) = e^textcolor{blue}{x} and v(x)=sin⁡xv(textcolor{blue}{x}) = sin textcolor{blue}{x}. Then

dydx=exsin⁡x−excos⁡xsin⁡2x=ex(sin⁡x−cos⁡x)sin⁡2x=ex(sin⁡x−cos⁡x)cosec⁡2xbegin{aligned}dfrac{dtextcolor{limegreen}{y}}{dtextcolor{blue}{x}} &= dfrac{e^textcolor{blue}{x}sin textcolor{blue}{x} – e^textcolor{blue}{x}cos textcolor{blue}{x}}{sin ^2 textcolor{blue}{x}}[1.2em]&=dfrac{e^textcolor{blue}{x}(sin textcolor{blue}{x} – cos textcolor{blue}{x})}{sin ^2 textcolor{blue}{x}}[1.2em]&=e^textcolor{blue}{x}(sin textcolor{blue}{x} – cos textcolor{blue}{x}) cosec^2 textcolor{blue}{x} end{aligned}

A LevelAQAEdexcelOCR

Example 2: Using the Quotient Rule (with the Product Rule)

Let y=(x2−1)ln⁡xcos⁡xtextcolor{limegreen}{y} = dfrac{(textcolor{blue}{x}^2 – 1)ln textcolor{blue}{x}}{cos textcolor{blue}{x}} where xtextcolor{blue}{x} is measured in radians. Find dydxdfrac{dtextcolor{limegreen}{y}}{dtextcolor{blue}{x}} and verify that there is a stationary point at the point (1,0)(1, 0).

[5 marks]

Let u(x)=(x2−1)ln⁡xu(textcolor{blue}{x}) = (textcolor{blue}{x}^2 – 1)ln textcolor{blue}{x} and v(x)=cos⁡xv(textcolor{blue}{x}) = cos textcolor{blue}{x}. Also, set a=x2−1a = textcolor{blue}{x}^2 – 1 and b=ln⁡xb = ln textcolor{blue}{x}. Then

dadx=2xdfrac{da}{dtextcolor{blue}{x}} = 2textcolor{blue}{x} and dbdx=1xdfrac{db}{dtextcolor{blue}{x}} = dfrac{1}{textcolor{blue}{x}}

so

du(x)dx=2xln⁡x+x2−1xdfrac{du(textcolor{blue}{x})}{dtextcolor{blue}{x}} = 2textcolor{blue}{x}ln textcolor{blue}{x} + dfrac{textcolor{blue}{x}^2 – 1}{textcolor{blue}{x}} and dv(x)dx=−sin⁡xdfrac{dv(textcolor{blue}{x})}{dtextcolor{blue}{x}} = -sin textcolor{blue}{x}

This gives

dydx=cos⁡x(2xln⁡x+x2−1x)+(sin⁡x(x2−1)ln⁡x)cos⁡2xdfrac{dtextcolor{limegreen}{y}}{dtextcolor{blue}{x}} = dfrac{cos textcolor{blue}{x}left( 2textcolor{blue}{x}ln textcolor{blue}{x} + dfrac{textcolor{blue}{x}^2 – 1}{textcolor{blue}{x}}right) + (sin textcolor{blue}{x} (textcolor{blue}{x}^2 – 1)ln textcolor{blue}{x})}{cos ^2 textcolor{blue}{x}}

When x=1textcolor{blue}{x} = 1,

dydx=cos⁡1((2×0)+0)+sin⁡1×0×0cos⁡21dfrac{dtextcolor{limegreen}{y}}{dtextcolor{blue}{x}} = dfrac{cos 1((2 times 0) + 0) + sin 1 times 0 times 0}{cos ^2 1}

=0= 0

So, we can confirm that there is a stationary point at x=1textcolor{blue}{x} = 1.

A LevelAQAEdexcelOCR

Quotient Rule Example Questions

Question 1: Using the quotient rule, show that the function y=1x3y = dfrac{1}{x^3} has derivative dydx=−3x4dfrac{dy}{dx} = dfrac{-3}{x^4}.

[3 marks]

A Level AQAEdexcelOCR

Let u(x)=1u(x) = 1 and v(x)=x3v(x) = x^3.

Then

dydx=0−3x2x6dfrac{dy}{dx} = dfrac{0 – 3x^2}{x^6}

 

=−3x4= dfrac{-3}{x^4}

MME Premium Laptop

Save your answers with

MME Premium

Gold Standard Education

Sign Up Now
Still marking manually? Save your answers with
MME Premium Find out more

Question 2: For f(x)=2sin⁡xcos⁡xf(x) = dfrac{2sin x}{cos x}, use the quotient rule to find its derivative with respect to xx, and prove that 2sin⁡xcos⁡x=2tan⁡xdfrac{2sin x}{cos x} = 2tan x.

[5 marks]

A Level AQAEdexcelOCR

Let u(x)=2sin⁡xu(x) = 2sin x and v(x)=cos⁡xv(x) = cos x. Then

 

df(x)dx=2cos⁡xcos⁡x+2sin⁡xsin⁡xcos⁡2xdfrac{df(x)}{dx} = dfrac{2cos x cos x + 2sin x sin x}{cos ^2 x}

 

=2sec⁡2x= 2sec ^2 x, by the identity sin⁡2x+cos⁡2x≡1sin ^2 x + cos ^2 x equiv 1.

We also have

d(2tan⁡x)dx=2sec⁡2xddfrac{(2tan x)}{dx} = 2sec ^2 x

Therefore, df(x)dx=dtan⁡xdxdfrac{df(x)}{dx} = dfrac{dtan x}{dx}

This is an example of the Uniqueness Theorem. (See Product Rule, Q2).

MME Premium Laptop

Save your answers with

MME Premium

Gold Standard Education

Sign Up Now
Still marking manually? Save your answers with
MME Premium Find out more

Question 3: Find the derivative (w.r.t xx), of the function y=x33xy = dfrac{x^3}{3^x}. Verify that there is a stationary point at the origin.

[5 marks]

A Level AQAEdexcelOCR

Set u=x3u = x^3 and v=3xv = 3^x.

Then

u′=3x2u' = 3x^2 and v′=3xln⁡3v' = 3^xln 3

Using the quotient rule, we have

dydx=3x+1x2−x33xln⁡332xdfrac{dy}{dx} = dfrac{3^{x + 1}x^2 – x^3 3^xln 3}{3^{2x}}

Set x=0x = 0 to give

 

dydx=(3×0)−(0×1×ln⁡3)1dfrac{dy}{dx} = dfrac{(3 times 0) – (0 times 1 times ln 3)}{1}

 

=0= 0

So, we can confirm there is a stationary point at the origin.

MME Premium Laptop

Save your answers with

MME Premium

Gold Standard Education

Sign Up Now
Still marking manually? Save your answers with
MME Premium Find out more

Additional Resources

Site Logo

Exam Tips Cheat Sheet

A Level
Open
Site Logo

Formula Booklet

A Level
Open

Specification Points Covered

G4 – Differentiate using the product rule, the quotient rule and the chain rule, including problems involving connected rates of change and inverse functions

https://www.google.com