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Math Help - Implicit Differentiation of a Trignometric Equation [Multiple Choice]

  1. #1
    Junior Member StarlitxSunshine's Avatar
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    Implicit Differentiation of a Trignometric Equation [Multiple Choice]

    If cos (xy) = x , then \frac{dy}{dx}=

    1. (A) \frac{-csc (xy) - y}{x}
    2. (B) \frac{csc (xy) - 1}{x}
    3. (C) -csc (xy)
    4. (D) \frac{-csc (xy)}{x}
    5. (A) -csc (xy) - 1
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by StarlitxSunshine View Post
    If cos (xy) = x , then \frac{dy}{dx}=

    1. (A) \frac{-csc (xy) - y}{x}
    2. (B) \frac{csc (xy) - 1}{x}
    3. (C) -csc (xy)
    4. (D) \frac{-csc (xy)}{x}
    5. (A) -csc (xy) - 1
    I give up...Which one is it?



    What's the problem, what don't you understand about the question? where are you stuck? Help me help you.
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  3. #3
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by StarlitxSunshine View Post
    If cos (xy) = x , then \frac{dy}{dx}=

    1. (A) \frac{-csc (xy) - y}{x}
    2. (B) \frac{csc (xy) - 1}{x}
    3. (C) -csc (xy)
    4. (D) \frac{-csc (xy)}{x}
    5. (A) -csc (xy) - 1
    Implicit differentiation gives
    -\sin(xy) (y + x y') = 1

    Solve for y':

    y' = \frac{\frac{-1}{sin(xy)}-y}{x}=\frac{-csc(xy)-y}{x}
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  4. #4
    Junior Member StarlitxSunshine's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    I give up...Which one is it?



    What's the problem, what don't you understand about the question? where are you stuck? Help me help you.
    >_< I know how you feel xD

    I tried implicit differentiation, but I don't think I'm doing it right. I've never done a question where the sin function has two variables (both x and y) in it, so I'm not really sure how that works. And moreover, I can't seem to get past the first step of the implicit differentiation because I don't know what to do next. :33
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by StarlitxSunshine View Post
    >_< I know how you feel xD

    I tried implicit differentiation, but I don't think I'm doing it right. I've never done a question where the sin function has two variables (both x and y) in it, so I'm not really sure how that works. And moreover, I can't seem to get past the first step of the implicit differentiation because I don't know what to do next. :33
    "implicit" means that it is "implied" that y is a function of x. This means that when x and y are grouped together as a product, quotient, etc... it must be understood that one must employ the neccesary and appropriate method of differetiating.

    For example, consider the function

    f(x)=y=x^2+x

    Taking the derivative is simple enough

    f'(x)=\frac{dy}{dx}=2x+1

    Well, what if we had simply rewrote before differentiating, and subtracted y from both sides

    0=x^2+x-y

    No problem, y is still a function of x, so the derivative "with respect to x" is

    0=2x+1-\frac{dy}{dx}

    And adding, we are back to where we started

    \frac{dy}{dx}=2x+1.

    In this case, it was very easy to solve for y, but sometimes - as in the problem you have provided - solving for y is tedious, and in some cases, impossible. So, we understand y to be an "implied" function of xand we differentiate.

    Bye!
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