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    How do you differentiate with respect to x" (x-1)^3 e^(x^2)-2x

    I get stuck because of the e^(x^2)-2x bit. :rolleyes:
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    (Original post by TheQueen1986)
    How do you differentiate with respect to x" (x-1)^3 e^(x^2)-2x

    I get stuck because of the e^(x^2)-2x bit. :rolleyes:
    Do you mean e^(x^2 - 2x)?

    If so: y = (x - 1)^3.e^(x^2 - 2x)

    u = (x - 1)^3, v = e^(x^2 - 2x)

    Applying the product rule: dy/dx = vdu/dx + udv/dx (and chain rule actually, cos you have a function of a function)

    dy/dx = [e^(x^2 - 2x)].3(x - 1)^2.1 + [(x - 1)^3].dv/dx

    v = e^(x^2 - 2x)

    Let t = x^2 - 2x, then v = e^t

    By the chain rule, dv/dx = (dv/dt)(dt/dx)

    dv/dx = e^t.(2x - 2)

    dv/dx = (2x - 2)e^(x^2 - 2x)

    => dy/dx = [e^(x^2 - 2x)].3(x - 1)^2.1 + [(x - 1)^3].(2x - 2)e^(x^2 - 2x)
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    Thanks I get that now. It was similar to the y=ln ... problem.

    The answer at the back wants it in this form: (x-1)^2 e(x^2-2x)[2x^2-4x+5]. Does it matter?
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    (Original post by TheQueen1986)
    Thanks I get that now. It was similar to the y=ln ... problem.

    The answer at the back wants it in this form: (x-1)^2 e(x^2-2x)[2x^2-4x+5]. Does it matter?
    Well, you could just put it in that form:

    => dy/dx = [e^(x^2 - 2x)].3(x - 1)^2.1 + [(x - 1)^3].(2x - 2)e^(x^2 - 2x)

    Take out common factors (ie. (x - 1)^2 and e^(x^2 - 2x)):

    dy/dx = [(x - 1)^2.e^(x^2 - 2x)][3 + (x - 1)(2x - 2)]

    = [(x - 1)^2.e^(x^2 - 2x)](3 + 2x^2 - 2x - 2x + 2)

    = [(x - 1)^2.e^(x^2 - 2x)](2x^2 - 4x + 5) as required.

    Not sure it matters too much, as both are correct.
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    the question is
    e^(xy)=4
    i tried a lot. is there anyone who can help? :
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    (Original post by bodomx)
    the question is
    e^xy=4
    i tried a lot. is there anyone who can help? :
    I’m not sure how to read your function.
    Is it
    (e^x)y = 4 ?
    or
    e^(xy)=4 ?

    If it’s (e^x)y = 4 then …

    Divide both sides by e^x, giving
    y = 4e^(-x)
    dy/dx = 4*(-1)*e^(-x)
    dy/dx = -4/(e^x)
    =============

    If it’s e^(xy) = 4 then …

    use substitution and the product rule.

    Let u = xy

    du/dx = x.dy/dx + y.(1)

    Then,

    e^u = 4

    differentiating both sides wrt x, using the chain rule,

    d(e^u)/dx = d(4)/dx
    d(e^u)/du * du/dx = 0
    e^u * (x.dy/dx + y) = 0
    x.dy/dx + y = 0
    dy/dx = -y/x
    =========
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    I got another one, here it is
    Find the value of sinθ when 7secθ-3tanθ is minimum.
    I did like this> first I differenciated 7secθ-3tanθ then equated it to zero. I couldn't get the answer this way. Then I used another method. I equated the 7secθ-3tanθ to zero. And then divided both sides by secθ. Got the book answer this way. but i think this makes no sense since the question says when it is at minimum. Anyone got an idea about this?
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    (Original post by bodomx)
    I got another one, here it is
    Find the value of sinθ when 7secθ-3tanθ is minimum.
    I did like this> first I differenciated 7secθ-3tanθ then equated it to zero. I couldn't get the answer this way. Then I used another method. I equated the 7secθ-3tanθ to zero. And then divided both sides by secθ. Got the book answer this way. but i think this makes no sense since the question says when it is at minimum. Anyone got an idea about this?
    erm - you need to rearrange the 7sec0-3tan0 (i dont have a theta key :rolleyes: ) into a simpler trig function to differentiate it - did you do that?
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    (Original post by bodomx)
    I got another one, here it is
    Find the value of sinθ when 7secθ-3tanθ is minimum.
    I did like this> first I differenciated 7secθ-3tanθ then equated it to zero. I couldn't get the answer this way. Then I used another method. I equated the 7secθ-3tanθ to zero. And then divided both sides by secθ. Got the book answer this way. but i think this makes no sense since the question says when it is at minimum. Anyone got an idea about this?
    If you equate 7secθ-3tanθ to zero, then you get,

    7secθ-3tanθ = 0
    7 - 3sinθ = 0
    sinθ = 7/3

    but sinθ is never greater than 1! So there's a mistake in doing it this way.

    My way!
    ======

    let f = 7secθ-3tanθ
    f' = 7secθ.tanθ - 3sec²θ = 0 for a turning point.
    7secθ.tanθ - 3sec²θ = 0
    7tanθ - 3secθ = 0
    7tanθ = 3secθ
    7sinθ = 3 - multiplying by cosθ
    sinθ = 3/7
    ========
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    (Original post by Fluffstar)
    erm - .... (i dont have a theta key :rolleyes: ) ...
    erm - you can enter θ into your posts by typing &#952 immediately followed by a semicolon.
    If you want any more symbols like that, search these forums for the text "special characters and symbols"
 
 
 
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