AQA Core 3 - Product rule

So I've got y = ln(3x+1)

I can't get the right answer but I don't know why? Essentially, I'll write my workings and the expected answer, if someone could point out what I'm doing it would be greatly appreciated!!

u = lnx
du/dx = 1/x
v = 3x+1
dv/dx = 3

dy/dx = u*dv/dx + v*du/dx

= lnx*3 + 3x+1*1/x

= 3lnx + 1/x (3x+1)

I'm supposed to get 3/(3x+1)
Have I not rearranged it or have I messed up? Thank you!

Reply 1

Smaug123

15

You've calculated

\frac{d}{dx}(\log(x) (3x+1))

, as far as I can tell. You need to be using the chain rule here, not the product rule (what you're trying to differentiate isn't a product).

Reply 2

m4ths/maths247

7

Original post by Meccaback

So I've got y = ln(3x+1)

I can't get the right answer but I don't know why? Essentially, I'll write my workings and the expected answer, if someone could point out what I'm doing it would be greatly appreciated!!

u = lnx
du/dx = 1/x
v = 3x+1
dv/dx = 3

dy/dx = u*dv/dx + v*du/dx

= lnx*3 + 3x+1*1/x

= 3lnx + 1/x (3x+1)

I'm supposed to get 3/(3x+1)
Have I not rearranged it or have I messed up? Thank you!

Take it back to the start, let u = 3x+1 which will give you ln(u)...now go ahead with the chain rule. :smile:

Reply 3

Meccaback

OP

Oh my God I'm so stupid. Tried again, using the chain rule this time... Worked. :L Thank you so much, couldn't work out what I was doing wrong!

Reply 4

TenOfThem

18

Original post by Meccaback

So I've got y = ln(3x+1)

There is a common mis-understanding amongst students that ln[f(x)] means ln multiplied by f(x)

This is nonsense

ln is a function so you have a function of a function

So, as has been said, chain rule

Reply 5

Hunarench95

13

Use the chain rule (function of a function). You can also think about it as a reverse integration, what would integrate to give that?

So when you integrate f'(x)/f(x) the result is ln(f(x)). Might help you a little bit.

AQA Core 3 - Product rule

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