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# Differentiating e^x using composite/chain rule. Watch

1. say for example I have to differentiate the function using composite rule (dy/dx=dy/du*du/dx):

y=(2x^2-1)^4

I call the bits in the bracket 'u', and 'y=(x^2-1)^4' becomes y=u^4, then I can differentiate both and apply the chain rule..

then take the function:

y=e^(2x^2-1)

.. Now I have a problem, I have an extra contant (K) of '2' before the x squared.

differenciation of e^kx = ke^kx

What on earth do I call the separate parts of y=e^(2x^2-1)?
I need a 'u' and a 'y', I assume.

Can someone please tell me/work this out with working?
tia.
2. Let u= 2x^2-1 and proceed as before.
3. You can just let u = 2x^2 - 1?
4. just call it v, w, z. Whatever as long as you don't use the same symbol to denote 2 different functions or variables
5. Call them whatever you want - you don't get in trouble for choosing your own letters for variables
6. (Original post by ghostwalker)
Let u= 2x^2-1 and proceed as before.

I'm sure the answer has a constant before the e^x
7. can someone do it and show working?

y=e^(2x^2-1)
8. y=e^(2x^2-1)
let u = (2x^2)-1
du/dx = 4x
y=e^u
dy/du = e^u
dy/dx = dy/du * du/dx
dy/dx = 4xe^u
dy/dx = 4xe^(2x^2-1)
9. Beaten to it, but oh well.

10. Thanks all.

Updated: June 16, 2009
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