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First Derivative

  1. Nov 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the first derivative for

    x sin(y) = y cos(x)

    3. The attempt at a solution
    x sin(y) = y cos(x)

    Product rule on each side:

    x cos(y) + sin(y) = -y sin(x) + cos(x)

    I'm not sure what to do after that or if if I was even doing it right to begin with.
     
  2. jcsd
  3. Nov 5, 2008 #2

    HallsofIvy

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    What are you differentiating with respect to- x or y? It must be one or the other. If you are differentiating with respect to x, then the derivative of sin(y) is NOT cos(y), it is, by the chain rule, cos(y)(dy/dx) and the derivative of y is NOT 1, it is dy/dx. If you are differentiating with respect to y, then the derivative of x is NOT 1, it is dx/dy, and the derivative of cos(x) is NOT -sin(x), it is -sin(x)(dx/dy).

    Finally, what does the problem really say? I am sure that it does not say "Find the first derivative for x sin(y) = y cos(x)", that doesn't make sense. I suspect it said "Find the derivatvive of y if x sin(y)= y cos(x)" though I can't be sure.
     
  4. Nov 5, 2008 #3
    This is all it says:

    "Find the first derivative for
    x sin(y) = y cos(x)"

    It does have answer choices, but they don't have any equal signs in them. They don't say "dy/dx =" or "dx/dy =". Here are the answer choices:

    A) Sin(x) - x sin(y) / y cos(x) - cos(y)

    B) cos(x) - sin(y) / sin(x) - cos(y)

    C) y sin(x) + sin(y) / cos(x) - x cos(y)

    D) sin(x) + sin(y) / cos(x) + cos(y)

    E) None of the above
     
  5. Nov 5, 2008 #4

    gabbagabbahey

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    I'm sure that they want you to find [itex]dy \over{dx}[/itex] by implicit differentiation.
     
  6. Nov 5, 2008 #5
    Working it out again for dy/dx:

    dx/dx Sin(y) + x cos(y)dy/dx = dy/dx cos(x) - y sin(x)dx/dx

    dx/dx just goes to 1:
    Sin(y) + x cos(y)dy/dx = dy/dx cos(x) - y sin(x)

    I'm pretty much where I was before except with a dy/dx on each side that would just cancel each other out.
     
  7. Nov 5, 2008 #6

    gabbagabbahey

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    The [itex]dy \over{dx} [/itex]'s don't "cancel each other out". There are two terms on each side of your equation, and only one term per side has [itex]dy \over{dx} [/itex] as a factor...Try isolating the terms with [itex]dy \over{dx} [/itex] on one side of the equation, and the rest of the terms on the other.
     
  8. Nov 5, 2008 #7
    Sin(y) + x cos(y)dy/dx + y sin(x) / cos(x)= dy/dx

    How would I get rid of the 2nd dy/dx? :confused:
     
  9. Nov 5, 2008 #8
    Well seeing as there's no dy/dx in the answer, I tried to do the following:

    dx/dx Sin(y) + x cos(y)dy/dx = dy/dx cos(x) - y sin(x)dx/dx

    dx/dx Sin(y) + x cos(y)dy/dx

    dy/dx = -sin(y)/x cos(y)


    dy/dx cos(x) - y sin(x)dx/dx

    dy/dx=y sin(x) / cos(x)

    -sin(y)/x cos(y) = y sin(x) / cos(x)

    Cross multiply to get:

    y sin(x)*x cos(y) / cos(x)*-sin(y)

    But it's still not an answer choice
     
  10. Nov 5, 2008 #9

    gabbagabbahey

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    Let's start again from here:

    [itex]dy \over{dx} [/itex] is a variable, and it happens to be the variable you are trying to solve for in the above equation.

    Start by moving the terms that have [itex]dy \over{dx} [/itex] in them to one side of the equation and move the terms that don't have [itex]dy \over{dx} [/itex] in them to the other side.

    For example, if I had the equation [itex]Ax+sin(y) \frac{dy}{dx}=By\frac{dy}{dx}-3cos(x) [/itex], I would solve it as follows:

    Step 1; isolate the terms with [itex]dy \over{dx} [/itex]:

    [tex]Ax+sin(y) \frac{dy}{dx}=By\frac{dy}{dx}-3cos(x) \Rightarrow sin(y) \frac{dy}{dx}-By\frac{dy}{dx}=-Ax-3cos(x)[/tex]

    Step 2; factor out a [itex]dy \over{dx} [/itex]:

    [tex]sin(y) \frac{dy}{dx}-By\frac{dy}{dx}=-Ax-3cos(x) \Rightarrow \frac{dy}{dx}(sin(y)-By)=-Ax-3cos(x) [/tex]

    Step 3: divide by [itex](sin(y)-By)[/itex] and hence solve for [itex]dy \over{dx} [/itex]:

    [tex]\frac{dy}{dx}=\frac{-Ax-3cos(x)}{(sin(y)-By)} [/tex]

    and so my answer would be [tex]\frac{-Ax-3cos(x)}{(sin(y)-By)}[/tex]

    Apply this method to your problem.
     
    Last edited: Nov 5, 2008
  11. Nov 5, 2008 #10
    y sin(x) + sin(y) / cos(x)-x cos(y)

    Wow, that was really easy. I was way over thinking that one :yuck:

    Thanks
     
  12. Nov 5, 2008 #11

    gabbagabbahey

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    Yep :approve:
     
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