The Student Room Group

Differentiating e^x using composite/chain rule.

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.
Let u= 2x^2-1 and proceed as before.
You can just let u = 2x^2 - 1?
Reply 3
just call it v, w, z. Whatever as long as you don't use the same symbol to denote 2 different functions or variables
Call them whatever you want - you don't get in trouble for choosing your own letters for variables
Reply 5
ghostwalker
Let u= 2x^2-1 and proceed as before.


What about e^x?

I'm sure the answer has a constant before the e^x
Reply 6
can someone do it and show working?

y=e^(2x^2-1)
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)
Beaten to it, but oh well.

[br]y=e2x21[br][br]u=2x21[br][br]y=eu[br][br] [br]y = e^{2x^2 - 1} [br][br]u = 2x^2 - 1[br][br]y = e^u[br][br]

dudx=4x\frac{du}{dx}=4x

dydu=eu\frac{dy}{du} = e^u

dydx=dydu×dudx\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}

dydx=(4x)e2x21\frac{dy}{dx} = (4x)e^{2x^2-1}
Reply 9
Thanks all.