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1. okay so if you want to look here's the question:
http://www.examsolutions.co.uk/a-lev...6/question.php

I'm stuck on b) where I have to differentiate 2^x^2 . HELP? I don't know what to do.

2. I really don't understand how it works.

u = x^2
du/dx = 2x

d/dx = du/dx * d/du

And you already found d/du in part a).
4. (Original post by AnonyMatt)

u = x^2
du/dx = 2x

d/dx = du/dx * d/du

And you already found d/du in part a).
I've never learnt the chain rule like this. So this is where I think I'm having problems.
5. (Original post by Josieee)
I've never learnt the chain rule like this. So this is where I think I'm having problems.
Take the definition of the chain rule to be the following:

dy/dx = dy/du * du/dx.

So if you have y = sin(x^2)

Let u = x^2, then y = sin(u), so dy/du = cos(u)

And u = x^2 so du/dx = 2x

By definition, dy/dx = dy/du * du/dx = cos(u) * 2x (as worked out above)

But u = x^2, so dy/dx = 2xcos(u) = 2xcos(x^2)

You can apply it here also.
6. Have you actually done part a)?

Someone above has shown it to be true through logarithmic differentiation, but this is the 'hence' method:

There's a full solution there, given that you already have the logarithmic one, but I have split it up into spoilers so you can try and get it yourself.

EDIT: I hate LaTeX so ******* much.

y = 2^{x} = e^{ln2^{x}} = e^{xln2}
u = xln2
Spoiler:
Show

du/dx = ln2
dy/du/ = e^{u}
Spoiler:
Show

dy/dx = e^{u} * ln2 = e^{xln2} * ln2 = 2^{x} * ln2
• (Original post by john1234567)

I think your supposed to sub in y afterwards

• okay I understand it now, thankyou
•
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