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Partial derivative of tan(x + y)

  1. Apr 12, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]f(x, y) = \tan(x + y) \\
    f_x = ?
    [/tex]



    2. Relevant equations

    [tex]\frac{dy}{dx}\tan(x)= \sec^2 x[/tex]

    3. The attempt at a solution

    I set y as constant, so I said derivative of y = 0 then took derivative of tan as above. However the answer should be
    [tex]f_x = \sec^2(x + y)[/tex]

    Why is y included?
     
  2. jcsd
  3. Apr 12, 2009 #2

    tiny-tim

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    Hi username12345! :smile:
    Because it's in g(x,y) in the chain rule … ∂f(g(x,y))/∂x = ∂f(g(x,y))/∂g ∂g(x,y))/∂x
     
  4. Apr 12, 2009 #3

    HallsofIvy

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    The fact that the derivative of y is 0 does not mean that y itself is 0 and replacing tan(x+y) with tan(x) is saying y= 0.

    What is the derivative of tan(x+ a) for constant a?
     
  5. Apr 12, 2009 #4
    Well, what if you asked, what is derivative of [tex]tan(2x + a)[/tex]... I would say [tex]2 \sec^2(2x + a)[/tex]. So if that is correct then the derivative of [tex]\tan(x + a)[/tex] would be [tex]\sec^2(x + a)[/tex]. Is this correct?

    Now considering the partial derivative with respect to x, we want y as a constant so replace a with y and we get [tex]f_x = \sec^2(x + y)[/tex]
     
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