x Turn on thread page Beta
 You are Here: Home >< Maths

# Core 3 Differentiation watch

1. Hi,
I know this is probably an easy question but how would you go about differentiating 4ln2x? Thanks
2. (Original post by student1856)
Hi,
I know this is probably an easy question but how would you go about differentiating 4ln2x? Thanks
Use the Chain rule.
3. (Original post by zetamcfc)
Use the Chain rule.
What part of it would I set as U? Would it be y=4U with U equalling ln2x?
4. (Original post by student1856)
What part of it would I set as U? Would it be y=4U with U equalling ln2x?
u=2x.
5. (Original post by student1856)
Hi,
I know this is probably an easy question but how would you go about differentiating 4ln2x? Thanks

6. 4 is a constant,
dy/dx(Inx) would be 1/x

as for ln2x:

dy/dx(In2x) = dy/dx(Inu) = 2/u = 2/2x = 1/x

so, dy/dx = constant/x.
7. (Original post by zetamcfc)
u=2x.
Right, so I do that and I get 2 multiplied by 4/x which gives me 8/x. But the answer is 4/x.
8. (Original post by student1856)
Right, so I do that and I get 2 multiplied by 4/x which gives me 8/x. But the answer is 4/x.
9. (Original post by student1856)
Right, so I do that and I get 2 multiplied by 4/x which gives me 8/x. But the answer is 4/x.
y=4lnu implies dy/dx = 4/u * du/dx
10. (Original post by XOR_)
4 is a constant,
dy/dx(Inx) would be 1/x

as for ln2x:

dy/dx(In2x) = dy/dx(Inu) = 2/u = 2/2x = 1/x

so, dy/dx = constant/x.

So, why is 4 a constant here but when you calculate 4lnx, 4 is not a constant? Why do you include the 4 in the diffeentiationof 4lnx but not in 4ln2x? Thank you
11. (Original post by student1856)
So, why is 4 a constant here but when you calculate 4lnx, 4 is not a constant? Why do you include the 4 in the diffeentiationof 4lnx but not in 4ln2x? Thank you
dy/dx(4lnx) = 4/x

dy/dx(4ln2x) = 4/x

for dy/dx(ln2x),
when u = 2x, you get dy/dx(lnu) = 2/2x = 1/x.
12. (Original post by student1856)
So, why is 4 a constant here but when you calculate 4lnx, 4 is not a constant? Why do you include the 4 in the diffeentiationof 4lnx but not in 4ln2x? Thank you
Would it help if we did this:

.

So:
13. (Original post by XOR_)
dy/dx(4lnx) = 4/x

dy/dx(4ln2x) = 4/x

for dy/dx(ln2x),
when u = 2x, you get dy/dx(lnu) = 2/2x = 1/x.
Ahh, I get it now. Thank you very much
14. (Original post by Zacken)
Would it help if we did this:

.

So:
Yes it does, thanks Zacken
15. (Original post by student1856)
Yes it does, thanks Zacken

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: May 15, 2016
Today on TSR

### Been caught plagiarising...

...for the 2nd time this year

### Mum says she'll curse me if I don't go to uni

Discussions on TSR

• Latest
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

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE