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Math Help - Implicit Differentiation Problems

  1. #1
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    Implicit Differentiation Problems

    4 problems I can't figure out:

    1) (x^2)(y^2) - y = x
    2) 2sinx(cosy) = 1
    3) sinx = x(1+tany)
    4) y = sin(xy)

    Please don't simplify or try to isolate y from the beginning. =)
    Only simplify to isolate dy/dx
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  2. #2
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    Quote Originally Posted by helpplz View Post
    4 problems I can't figure out:

    1) (x^2)(y^2) - y = x

    product rule with the \textcolor{red}{x^2y^2} term

    2) 2sinx(cosy) = 1

    product rule again ...

    3) sinx = x(1+tany)

    product rule on the right ...

    4) y = sin(xy)

    product rule with the xy argument
    what seems to be your problem with using the product rule for derivatives?
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  3. #3
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    I'm not sure why I just don't know where to start..I think I just don't know how to apply product rule with the y's..
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  4. #4
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    Quote Originally Posted by helpplz View Post
    I'm not sure why I just don't know where to start..I think I just don't know how to apply product rule with the y's..
    (1st function)(derivative of the 2nd) + (2nd function)(derivative of the 1st)

    like so ...

    \frac{d}{dx}(x^2y^2) = x^2 \cdot 2y \frac{dy}{dx} + y^2 \cdot 2x
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    Hm, but why is it 2y(dy/dx) instead of just 2(dy/dx)?
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  6. #6
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    Quote Originally Posted by helpplz View Post
    Hm, but why is it 2y(dy/dx) instead of just 2(dy/dx)?
    because it was y^2 to begin with ... not 2y. you still have to use the power rule.
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  7. #7
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    Quote Originally Posted by skeeter View Post
    because it was y^2 to begin with ... not 2y. you still have to use the power rule.
    So if it was something like y^3, the derivative would be 3y(dy/dx)^2 ?
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  8. #8
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    Quote Originally Posted by helpplz View Post
    So if it was something like y^3, the derivative would be 3y(dy/dx)^2 ?
    no

    \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}
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  9. #9
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    Quote Originally Posted by skeeter View Post
    no

    \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}
    Ah, okay, gotcha. I'm gonna do a random problem just to make sure I'm doing it right:

    x^2y^2 -2x = 3

    x^2(2y(dy/dx)) + y^2(2x) = 5

    dy/dx = 5-y^2(2x)/ x^2(2y)

    That right?
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  10. #10
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    Quote Originally Posted by helpplz View Post
    Ah, okay, gotcha. I'm gonna do a random problem just to make sure I'm doing it right:

    x^2y^2 -2x = 3

    x^2(2y(dy/dx)) + y^2(2x) = 5

    not quite ...

    \textcolor{red}{x^2 \cdot 2y \frac{dy}{dx} + y^2 \cdot 2x - 2 = 0}
    the derivative of a constant is 0.
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  11. #11
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    Quote Originally Posted by skeeter View Post
    the derivative of a constant is 0.
    Oh geez, I must have forgotten that on at least 15 problems...Thanks sooo much for correcting me. =) And for all your help!
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