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Math Help - Trig derivatives

  1. #16
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    Quote Originally Posted by rmn View Post
    ok so I think this is using chain rule again but i'm stuck at simplifying

    y= ((1+cosx)/(sinx))^(-1)

    i'm at

    -1 (1+cosx/sinx) ^(-2) times (-sinx^2 - cos x +cos x^2)/sinx^2

    and also

    y=sec^(2) pi x
    ok so there isn't any brackets around pix its just right beside sec^2 so i dont know how the chain rule would work?
    Here is one way.
    Sorry but I forgot already those chain rule, quotient rule, whatever rule. I just keep on differentiating until all the variables are differentiated, etc. I must be following those said rules although those rules are not in my mind any more.

    y = [(1 +cosX) / sinX]^(-1)

    dy/dX = -1[(1 +cosX) / sinX]^(-2) *({sinX[-sinX] -(1 +cosX)[cosX]} / sin^2(X))

    dy/dX = (-1 / [(1 +cosX) / sinX]^2) * ({-sin^2(X) -cosX -cos^2(X)} / sin^2(X))

    dy/dX = -(sin^2(X) / (1 +cosX)^2) * -({sin^2(X) +cos^2(X) +cosX} / sin^2(X))

    The sin^2(X) cancell out; the sin^2(X) +cos^2(X) is converted to 1; the multiplication of negatives turns to positive,

    dy/dX = (1 +cosX) / (1 +cosX)^2

    dy/dX = 1 / (1 +cosX) ------------------------answer.

    ---------------------------------------------------------------

    y = sec^2(pi*X)
    Or,
    y = [sec(pi*x)]^2


    dy/dX = 2[sec(pi*X)] *[sec(pi*X)tan(pi*X)] *[pi]

    dy/dX = (2pi)sec^2(pi*X)tan(pi*X) -----------------answer.
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  2. #17
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    same

    Iv also got a problem with derivative of trig functions, heres mine:

    [sinInverse(x)/sqrt(1-x^2) dx] for range 0...0.5

    am not sure how to solve that?
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  3. #18
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by taurus View Post
    Iv also got a problem with derivative of trig functions, heres mine:

    [sinInverse(x)/sqrt(1-x^2) dx] for range 0...0.5

    am not sure how to solve that?
    are you sure you mean derivative? you have a dx attached to it. that means you should integrate
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  4. #19
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    sorry yea i mean integrate?
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  5. #20
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    \int \frac {\arcsin x}{\sqrt{1 - x^2}}~dx

    note that \frac d{dx} \arcsin x = \frac 1{\sqrt {1 - x^2}}

    thus, a simple substitution u = \arcsin x will do the trick here.

    try it out.


    EDIT: whatever, i'll just do it

    Let u = \arcsin x

    \Rightarrow du = \frac 1{\sqrt {1 - x^2}}~dx

    So our integral becomes

    \int u ~du = \frac 12u^2 + C = \frac 12 \arcsin^2 x + C
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  6. #21
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    just a quick question, where did the half come from?
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  7. #22
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by taurus View Post
    just a quick question, where did the half come from?
    it is from the power rule for integrals. \int x^n~dx = \frac {x^{n + 1}}{n + 1} + C for n \ne -1

    thus \int u~du = \frac {u^{1 + 1}}{1 + 1} + C = \frac 12u^2 + C
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