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# C3/C4 Differentiation - Help needed Watch

1. x=t^2e^t

I realise you have to use the product rule:

v (du/dx) + u (dv/dx)

Let t^2=u

e^t=v

du/dx = 2t

dv/dx = 0 surely?

Textbook has: (t^2)(e^t) + (e^t)2t

I do not understand the bold bit and where it comes from. Can someone help?
2. you have dv/dt = 0

but d(e^t)/dt = e^t

if you then multiply this by u you get (t^2)(e^t)

edit: as we need dv/dt * u + du/dt * v and we have replaced dv/dt = 0 with dv/dt = e^t and that's where the bold expression comes from
3. Wait, but aren't expressions like e^2, e^3 constants. Why doesn't t follow suit?
4. The differential of e^x is e^x, use chain rule to prove this to yourself

Spoiler:
Show
y=e^x
let y=e^u
u=x
dy/du=e^u
du/dx=1
dy/dx=e^x.1
=e^x
5. (Original post by Deep456)
Wait, but aren't expressions like e^2, e^3 constants. Why doesn't t follow suit?
t is a variable, and we are differentiating with respect to t, so wherever we see t we need to treat it as if it were x if we were differentiating with respect to x
6. (Original post by Deep456)
Wait, but aren't expressions like e^2, e^3 constants. Why doesn't t follow suit?
That's the same as asking why 2^x doesn't follow suit as 2^2 is a constant.
7. And why x^2 doesn't follow suit despite 4, 9 and 16 being constants.
8. (Original post by porkstein)
..
Do you want your challenging question solved or do you just have it there for fun lol
9. Oh ok, thanks.

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