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Math Help - Implicit differentiation with trig

  1. #1
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    Implicit differentiation with trig

    1. sin(x+y) = x-y The answer is (1-cos(x+y))/((cos x+y)+1)
    derivitive: cos(x+y)(1+dy/dx)=1-dy/dx
    (dy/dx)+cos(x+y)(1+dy/dx)=1
    (dy/dx)+cos(x+y)+cos(x+y(dy/dx))=1
    (dy/dx)+cos(x+y(dy/dx))=1-cos(x+y)
    I believe this is when i went wrong or something.

    2. cos(xy)=1-x^2 the answer is (2x-y sin (xy))/(x sin (xy))
    derivitive: -sin(xy)(x(dy/dx)+y)=-2x
    2x-sin(xy)(x(dy/dx)+y)=0
    And I got lost from here.

    3.ln(xy)=e^2x The answer is (2e^2x-x^-1)y
    derivitive: (1/xy)(xy'+y)=2e^2x
    (-xy'+x-xy'+y)/(xy)^2=2e^2x
    And then this step was when I got lost as to how this translates to the answer.
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  2. #2
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    Don't forget the chain rule with these.


    #1: sin(x+y)-x+y=0

    cos(x+y)(1+\frac{dy}{dx})-1+\frac{dy}{dx}=0

    cos(x+y)+\frac{dy}{dx}cos(x+y)-1+\frac{dy}{dx}=0

    \frac{dy}{dx}=\frac{1-cos(x+y)}{cos(x+y)+1}





    #2: cos(xy)-x^{2}=0

    (x\frac{dy}{dx}+y)(-sin(xy))+2x=0

    -xsin(xy)\frac{dy}{dx}-ysin(xy)+2x=0

    \frac{dy}{dx}=\frac{2x-ysin(xy)}{xsin(xy)}


    The last one.

    ln(xy)-e^{2x}=0

    \frac{x\frac{dx}{dy}+y}{xy}=2e^{2x}

    \frac{dy}{dx}=\frac{(2xe^{2x}-1)y}{x}

    The best way to do these is the way earboth showed.

    Take the partials with respect to x and y:

    \frac{d}{dx}[ln(xy)-e^{2x}]=\frac{1}{x}-2e^{2x}

    \frac{d}{dy}[ln(xy)-e^{2x}]=\frac{1}{y}

    -\frac{\frac{1}{x}-2e^{2x}}{\frac{1}{y}}=\frac{(2xe^{2x}-1)y}{x}

    Same as above. See?.
    Last edited by galactus; July 20th 2007 at 06:37 PM.
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  3. #3
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    I get 1 and 2. But for 3, the problem i have is how
    \frac{x\frac{dx}{dy}+y}{xy}=2e^{2x}

    Becomes this:

    <br />
\frac{dy}{dx}=\frac{(2xe^{2x}-1)y}{x}<br />

    Other than that, I get the rest of the material.
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  4. #4
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    Quote Originally Posted by driver327 View Post
    I get 1 and 2. But for 3, the problem i have is how
    \frac{x\frac{dx}{dy}+y}{xy}=2e^{2x}

    Becomes this:

    <br />
\frac{dy}{dx}=\frac{(2xe^{2x}-1)y}{x}<br />

    Other than that, I get the rest of the material.
    Hello,

    I'll show you the steps of transformation only:

    \frac{x\frac{dx}{dy}+y}{xy}=2e^{2x} . Your equation. Multiply by (xy)

    x\frac{dx}{dy}+y=2e^{2x}\cdot xy . Subtract y

    x\frac{dx}{dy}=2e^{2x}\cdot xy - y . Divide by x

    \frac{dx}{dy}=\frac{2e^{2x}\cdot xy - y}{x} . Factor the numerator:

    \frac{dx}{dy}=\frac{\left(2e^{2x}\cdot x - 1\right) y}{x}
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