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# ln Core 3 question Watch

1. in question 8c:

I get the exact same answer except for not getting the "x 1/2"

I dont get how they get that. when you differentiate ln , it goes to 1/x and not k(1/x) ?

http://store.aqa.org.uk/qual/gceasa/...W-QP-JUN06.PDF
http://store.aqa.org.uk/qual/gceasa/...W-MS-JUN06.PDF
2. You might to want to tell us the actual question...
3. bump
4. (Original post by uxa595)
bump
Just use chain rule surely? Make a u substitution. ln u differentiated becomes 1/u. And then u differentiated becomes 1/2. So its 1/2 x u = 1/(2u).
5. I realise they have used the chain rule to multiple the dy/dx of the inside of the brackets by the outsid..

But why have they used chain over product ( as we are multiplying 2 things right)?

Edit: posted this without refreshing, yep, i noticed they use the chain rule but i dont get why.

They are multiplying one thing by another, and you should use the product rule?
6. (Original post by uxa595)
I realise they have used the chain rule to multiple the dy/dx of the inside of the brackets by the outsid..

But why have they used chain over product ( as we are multiplying 2 things right)?

Edit: posted this without refreshing, yep, i noticed they use the chain rule but i dont get why.

They are multiplying one thing by another, and you should use the product rule?
Its the chain rule because its a function of a function. The 1/2 is just a scalar value multiplying lnu.
i.e In C1 when you differentiated 3x^2 you don't care about the 3. It just becomes 6x. (Because you are differentiating with respect to x).

Whereas if you have xlnu then you would have to use product rule because there are two terms including x (assuming u has some x terms).

So basically if there are two terms including x being multiplied together then you use product rule. And if its a function of a function including an x term, then you use chain rule.
7. hmm well i was looking at it as : 1/3 . ln(u) so it would be product rule.
But this is obviously not true in this case.

As far as i have learnt, if the function includes ln, e^, trig (sin,cos etc.) , you use chain rule?
8. (Original post by uxa595)
hmm well i was looking at it as : 1/3 . ln(u) so it would be product rule.
But this is obviously not true in this case.

As far as i have learnt, if the function includes ln, e^, trig (sin,cos etc.) , you use chain rule?
It totally depends on how the problem is set up.

These would all be chain rule:
y = e^2x
y = sin 4x
y = ln 3x^3
y = (3x^2 + 4)^2

since you can put them in the form
y = e^u
y = sin u
y = ln u
y = u^2

You would then differentiate y with respect to u

dy/du = e^u
dy/du = cos u
dy/du = 1/u
dy/du = 2u

Then you differentiate u with respect to x

du/dx = 2
du/dx = 4
du/dx = 9x^2
du/dx = 6x

Then you multiply dy/du by du/dx to get dy/dx and substitute u for your expression in terms of x

dy/dx = 2 x e^u = 2e^2x
dy/dx = 4 x cos u = 4cos 4x
dy/dx = (9x^2) x 1/(3x^3) = (9x^2)/(3x^3)
dy/dx = 2(u) x (6x)= 2(3x^2 + 4) x (6x) = (2)(3x^2 + 4x)(6x)

For product rule:
y = 2xe^x
y = xln2x

Well we have 2x multiply e^x and x multiply ln2x

So we can say let u = 2x and v = e^x
Then y =uv
So dy/dx = (u x dy/dv) + (v x dy/du) = (2 x e^x) + ((2x) x e^x)
and for next one
dy/dx = (u x dy/dv) + (v x dy/du) = ( (x) x 2/2x ) + (1 x ln2x) = (1 + ln2x)

However if it was for example
6e^x
or (1/3)ln2x
We use chain rule. Since we don't have two terms being multiplying each other that include x.
So dy/dx = dy/du x du/dx = 6e^u x 1 = 6e^u = 6e^x
and for the next one
dy/dx = dy/du x du/dx = 1/3u x 2 = 2/3(2x) = 1/3x

I hope this is useful.

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