You are Here: Home >< Maths

# Differentiate this equation please watch

1. Please can any maths people try to differentiate the following equation for me: y = -1/(x^4-8), where ^ means 'to the power'. Thanks a lot.
2. (Original post by play_the_world)
Please can any maths people try to differentiate the following equation for me: y = -1/(x^4-8), where ^ means 'to the power'. Thanks a lot.
By the quotient rule
y = u/v => dy/dx = [v.du/dx - u.dv/dx]/v²

u = -1 => du/dx = 0
v = [x4 - 8] => dv/dx = 4x³

dy/dx
= [(x4 - 8) x 0 - (-1 x 4x³)]/(x4 - 8)²
= (0 + 4x3)/(x4 - 8)²
= 4x³/(x4 - 8)²

By the chain rule
Let u = x4 - 8 => du/dx = 4x³
=> y = -1/u => dy/du = 1/u²

dy/dx
= dy/du x du/dx
= 1/u² x 4x³
= 4x³/u²
= 4x³/(x4 - 8)²
3. (Original post by Widowmaker)
By the quotient rule
y = u/v => dy/dx = [v.du/dx - u.dv/dx]/v²

u = -1 => du/dx = 0
v = [x4 - 8] => dv/dx = 4x³

dy/dx
= [(x4 - 8) x 0 - (-1 x 4x³)]/(x4 - 8)²
= (0 + 4x3)/(x4 - 8)²
= 4x³/(x4 - 8)²

By the chain rule
Let u = x4 - 8 => du/dx = 4x³
=> y = -1/u => dy/du = 1/u²

dy/dx
= dy/du x du/dx
= 1/u² x 4x³
= 4x³/u²
= 4x³/(x4 - 8)²
Thankyou for confirming the answer. That's what i got too! And such a quick response x x x

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: July 13, 2005
Today on TSR

### TSR Pub Quiz 2018 - Anime

The first of our week-long series of entertainment quizzes

### University open days

Wed, 21 Nov '18
• Buckinghamshire New University
Wed, 21 Nov '18
• Heriot-Watt University
Wed, 21 Nov '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams