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Math Help - Calculus Problems

  1. #1
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    Calculus Problems

    1. if \frac{dy}{dx}=4y and if y=4 when x=0, then y=

    2. \int_0^{1/2} \frac{2x dx}{\sqrt{1-x^2}}=

    3. the point on the curve 2y=x^2 nearest to (4,1) is

    4. if tan(xy) = x, then \frac{dy}{dx}=
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by yamuda View Post
    1. if \frac{dy}{dx}=4y and if y=4 when x=0, then y=

    2. \int_0^{1/2} \frac{2x dx}{\sqrt{1-x^2}}=

    3. the point on the curve 2y=x^2 nearest to (4,1) is

    4. if tan(xy) = x, then \frac{dy}{dx}=
    First one is an SDE so you go \frac{dy}{y}=4dx integrating we get ln|y|=4x+c then you get y=Ce^{4x}

    2. The derivative of the quantity is present therefore -\int_0^{\frac{1}{2}}\frac{2x}{\sqrt{1-x^2}}dx=-2\sqrt{1-x^2}|_0^{\frac{1}{2}}

    for the third one I think you can figure that out

    4.differenating implicitly we get sec^2(xy)\cdot{(xy'+y)}=1 now solve for y'
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