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Differentiating e^x using composite/chain rule.

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    ReputationRep:
    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. Offline

    Let u= 2x^2-1 and proceed as before.
  3. Offline

    ReputationRep:
    You can just let u = 2x^2 - 1?
  4. Offline

    ReputationRep:
    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. Offline

    ReputationRep:
    Call them whatever you want - you don't get in trouble for choosing your own letters for variables
  6. Offline

    ReputationRep:
    (Original post by 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
  7. Offline

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    can someone do it and show working?

    y=e^(2x^2-1)
  8. Offline

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    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. Offline

    ReputationRep:
    Beaten to it, but oh well.

     

y = e^{2x^2 - 1} 



u = 2x^2 - 1



y = e^u

    \frac{du}{dx}=4x

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

    \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}

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

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