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# C3 differentiation question watch

1. Differentiate 3e^x -0.5ln(x-2)

I know how to differentiate exponentials and natural logs
but i cant get the right answer.
I dont know what to do with the negative 0.5 in front of the natural log.
2. Bring the -0.5 to the front of the expression

and differentiate that using product rule...

EDIT

is the starting expression this

or

If its the second, then as Pheylan suggests...

3. can you do it now?
4. how does that work?
how can you take the -0.5 to the front?
why would it then be multiplied by the ln(x-2) shouldnt it be added
5. I don't think the product rule is necessary.
first differentiate 3x^x which gives us 3e^x
then differentiate -0.5ln(x-2) which will give us (-0.5/(x-2)) or rather : -1/(2x-4)
6. (Original post by New Username)
I don't think the product rule is necessary.
first differentiate 3x^x which gives us 3e^x
then differentiate -0.5ln(x-2) which will give us (-0.5/(x-2))

thats what i did and its wrong
7. (Original post by ks1234567)
thats what i did and its wrong
Bring the -0.5 to the front of the expression

and differentiate that using product rule...

EDIT

is the starting expression this

or

If its the second, then as Pheylan suggests...
it's the second
9. (Original post by New Username)
3e^x - 1/2x
10. (Original post by Pheylan)

can you do it now?
how do you get to that?
11. (Original post by ks1234567)
how do you get to that?
both terms have a factor of 0.5, so factorise it
12. (Original post by Pheylan)
both terms have a factor of 0.5, so factorise it
still a bit stuck though
so far this is what i've got

0.5(6e^x - 1/x-2)

0.5 * 1/x-2 doesnt give -1/2x which is the answer
13. (Original post by Pheylan)

can you do it now?

Don't mean to hijack this thread, but since I've got a maths exam on wed I thought I'd ask...
From the starting equation

Did you divide both sides by 1/2 to get ?

Spoiler:
Show
(Which would yield the answer )

Is that right?
14. (Original post by ks1234567)
still a bit stuck though
so far this is what i've got

0.5(6e^x - 1/x-2)

0.5 * 1/x-2 doesnt give -1/2x which is the answer
Your book is wrong, as you rightly said, it should be
15. (Original post by blueJ-)
Your book is wrong, as you rightly said, it should be
yeah thats what i've got
16. (Original post by ks1234567)
still a bit stuck though
so far this is what i've got

0.5(6e^x - 1/x-2)

0.5 * 1/x-2 doesnt give -1/2x which is the answer
hm..is this from a text book?

the standard result is that:

Perhaps they are using the first rule giving them 1/x and then multiplying that by 1/2 which would explain the 1/2x - however that answer isn't valid
17. (Original post by blueJ-)
Your book is wrong, as you rightly said, it should be
its an edexcel mark scheme
C3 june 2005
18. (Original post by ks1234567)
its an edexcel mark scheme
C3 june 2005
I doubt it, are you sure it's not a school-produced mark scheme?

If that's what it does say, it's not the latest version, because that's an error.
19. (Original post by ks1234567)
its an edexcel mark scheme
C3 june 2005
If it's june 2005, then the question is actually meant to be:

The -2 is not in the brackets!!
20. (Original post by .:excel4100%:.)
If it's june 2005, then the question is actually meant to be:

The -2 is not in the brackets!!
ahh that clears up things.. thanks a lot mate

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