Hey there Sign in to join this conversationNew here? Join for free

Differentiating e^x using composite/chain rule.

Announcements Posted on
    • Thread Starter
    • 0 followers
    Offline

    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.
    • 29 followers
    Offline

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

    ReputationRep:
    You can just let u = 2x^2 - 1?
    • 0 followers
    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
    • 0 followers
    Offline

    ReputationRep:
    Call them whatever you want - you don't get in trouble for choosing your own letters for variables
    • Thread Starter
    • 0 followers
    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
    • Thread Starter
    • 0 followers
    Offline

    ReputationRep:
    can someone do it and show working?

    y=e^(2x^2-1)
    • 16 followers
    Offline

    ReputationRep:
    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)
    • 0 followers
    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}
    • Thread Starter
    • 0 followers
    Offline

    ReputationRep:
    Thanks all.

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?

    this is what you'll be called on TSR

  2. this can't be left blank
    this email is already registered. Forgotten your password?

    never shared and never spammed

  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide the button to the right to create your account

    Slide to join now Processing…

Updated: June 16, 2009
New on TSR

Naughtiest thing you did at school

Did you get away with it or were you punished?

Article updates
Reputation gems:
You get these gems as you gain rep from other members for making good contributions and giving helpful advice.