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Math Help - [SOLVED] Several questions

  1. #1
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    [SOLVED] Several questions

    1) How do you integrate Sin^2 x or cos^2 x I know that the integral of sin x is -cos x and that the integral of cos x is sin x but I don't know what to do with the square.

    2)How do you integrate e^5x I know that the integral of e^x is e^x so is it just e^5x?

    3)How do you do this: Find the tangent line to: (x^4 + 6xy +y^2 = 29) I don't know what to do with the y's
    Last edited by Amberosia32; February 4th 2010 at 03:42 PM. Reason: forgot a question
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  2. #2
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    Dear Amberosia32,

    For the first and second problems you should use the chain rule. That is if y and z are functons of x, \frac{dy}{dx}=\frac{dy}{dz}\times\frac{dz}{dx}.

    eg: \frac{d(sin^{2}x)}{dx}=\frac{d(sinx)^2}{d(sinx)}\t  imes{\frac{(sinx)}{dx}}=2sinxcosx=sin2x

    Similar method could be used to differentiate cos^{2}x and e^{5x}

    For a curve in two dimention(including a independent and dependent variable.) \frac{dy}{dx} gives the tangent of the curve.
    Therefore you could differentiate the equation, and afterwards subject \frac{dy}{dx},

    x^{4} + 6xy +y^{2} = 29

    4x^{3}+6x\frac{dy}{dx}+6y+2y\frac{dy}{dx}=0

    Subject \frac{dy}{dx} and you would get,

    \frac{dy}{dx}=-\frac{(2x^{3}+3y)}{3x+y}

    Here I have used the Leibniz's theorem in addition to the chain rule,

    That is if v and u are two functions of x, \frac{d(vu)}{dx}=v\frac{du}{dx}+u\frac{dv}{dx}

    Hope these things will help you to understand differentiation. But if you have any questions please don't hesitate to reply me.
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  3. #3
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    Quote Originally Posted by Sudharaka View Post
    Dear Amberosia32,

    For the first and second problems you should use the chain rule. That is if y and z are functons of x, \frac{dy}{dx}=\frac{dy}{dz}\times\frac{dz}{dx}.

    eg: \frac{d(sin^{2}x)}{dx}=\frac{d(sinx)^2}{d(sinx)}\t  imes{\frac{(sinx)}{dx}}=2sinxcosx=sin2x

    Similar method could be used to differentiate cos^{2}x and e^{5x}

    For a curve in two dimention(including a independent and dependent variable.) \frac{dy}{dx} gives the tangent of the curve.
    Therefore you could differentiate the equation, and afterwards subject \frac{dy}{dx},

    x^{4} + 6xy +y^{2} = 29

    4x^{3}+6x\frac{dy}{dx}+6y+2y\frac{dy}{dx}=0

    Subject \frac{dy}{dx} and you would get,

    \frac{dy}{dx}=-\frac{(2x^{3}+3y)}{3x+y}

    Here I have used the Leibniz's theorem in addition to the chain rule,

    That is if v and u are two functions of x, \frac{d(vu)}{dx}=v\frac{du}{dx}+u\frac{dv}{dx}

    Hope these things will help you to understand differentiation. But if you have any questions please don't hesitate to reply me.
    I don't understand how you got that 2sinxcosx=sin2x And in the other one were does the 6x or the 6y come from?
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  4. #4
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    Quote Originally Posted by Amberosia32 View Post
    I don't understand how you got that 2sinxcosx=sin2x
    This is a standard trig identity, you should commit it to memory

    Quote Originally Posted by Amberosia32 View Post
    I And in the other one were does the 6x or the 6y come from?
    Through implicit differentiation. Are you aware of this?
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  5. #5
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    Quote Originally Posted by pickslides View Post
    This is a standard trig identity, you should commit it to memory



    Through implicit differentiation. Are you aware of this?
    never heard of it....My class is more of a figure it out your self setting. But thanks.
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