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    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?
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    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
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    Wait, but aren't expressions like e^2, e^3 constants. Why doesn't t follow suit?
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    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
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    (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
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    (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.
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    And why x^2 doesn't follow suit despite 4, 9 and 16 being constants.
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    (Original post by porkstein)
    ..
    Do you want your challenging question solved or do you just have it there for fun lol
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    Oh ok, thanks.
 
 
 
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