You are Here: Home >< Maths

# C3 Chain Rule Watch

1. Came across this question and my solution is different from the textbook answer.
The curve C is defined by the equation:

Show that there is only one stationary point and find its coordinates.

I used chain rule then found dy/dx to be:

When I set this equal to 0 I proved there was one stationary point but my x-coordinate was 1/3 when the book had 2/3.

Where have I went wrong?
2. (Original post by ozmo19)
Came across this question and my solution is different from the textbook answer.
The curve C is defined by the equation:

Show that there is only one stationary point and find its coordinates.

I used chain rule then found dy/dx to be:

When I set this equal to 0 I proved there was one stationary point but my x-coordinate was 1/3 when the book had 2/3.

Where have I went wrong?

1/( 1 - x ) 3 = 8/(x)3

cube root both sides... 1/( 1 - x ) = 2/(x)

*now rearrange & solve to get the book answer*
3. (Original post by the bear)

1/( 1 - x ) 3 = 8/(x)3

cube root both sides... 1/( 1 - x ) = 2/(x)

*now rearrange & solve to get the book answer*
1/( 1 - x ) 3 = 1/8(x)3 by accident.

Thank you!
4. (Original post by ozmo19)
1/( 1 - x ) 3 = 1/8(x)3 by accident.

Thank you!
we are here to help *
5. (Original post by the bear)
we are here to help *

{\rtf1\ansi\ansicpg1252
{\fonttbl\f0\fnil\fcharset0 ArialMT;}{\colortbl;\red255\gree n255\blue255;\red80\green80\blue 80;}\deftab720\pard\pardeftab720 \partightenfactor0\f0\fs32 \cf2 \expnd0\expndtw0\kerning0\outl0\ strokewidth0 \strokec2 }
Attached Images

6. A: it is a product rule with a little bit of chain too...

u = e-x

v = 4e2x - 2ex

remember eax differentiates to aeax...

B: alternatively you could multiply out the brackets then you would not need the product rule ...
7. (Original post by the bear)
A: it is a product rule with a little bit of chain too...

u = e-x

v = 4e2x - 2ex

remember eax differentiates to aeax...

B: alternatively you could multiply out the brackets then you would not need the product rule ...
Took du/dx as e^-x and not -e^-x - Thanks for the reminder

Updated: October 16, 2016
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:
Today on TSR

### Stuck for things to do this summer?

Come and get some inspiration.

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