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Math Help - differentiating with multiple variables

  1. #1
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    differentiating with multiple variables

     h(z) = 3cos(z) + 3^x + cos^{-1}(2x)

     h'(z) = (3cos(z))' +  (3^x)' + (cos^{-1}(2x))'

     h'(z) = -3sin(z) + (3^x)' +  (cos^{-1}(2x))' //how do I finish?
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  2. #2
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    Re: differentiating with multiple variables

    If x and z are independent variables, then the function should be written as h(x,z), and the derivative with respect to z should be denoted by \frac{\partial h}{\partial z}. In this case, x is treated a constant, so \frac{\partial}{\partial z}(3^x + cos^{-1}(2x))=0.
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  3. #3
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    Re: differentiating with multiple variables

    Thanks/
    So variables are constants, unless they are taken with respect to the function?
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    Re: differentiating with multiple variables

    Quote Originally Posted by rabert1 View Post
    So variables are constants, unless they are taken with respect to the function?
    Sorry, this sentence does not make much sense.
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    Re: differentiating with multiple variables

    Quote Originally Posted by rabert1 View Post
    So variables are constants, unless they are taken with respect to the function?
    Have you studied implicit differentiation?

    Like a circle: x^2+y^2=r^2. We want to find y'=\frac{dy}{dx}.

    Thus 2x+2yy'=0 solve for y'.
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  6. #6
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    Re: differentiating with multiple variables

    thanks for replying!

    How would one know when the variable is a constant or not?
    See you have from the equation:

    (y^2)' = 2yy'

    and

    (r^2)' = 0
    Why wouldn't (r^2)' = 2rr'
    Last edited by rabert1; April 7th 2012 at 07:00 PM.
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  7. #7
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    Re: differentiating with multiple variables

    Quote Originally Posted by rabert1 View Post
    thanks for replying!
    How would one know when the variable is a constant or not?
    See you have from the equation: (y^2)' = 2yy'
    and
    (r^2)' = 0
    Why wouldn't (r^2)' = 2rr'
    One is expected to know that a circle is in two variables, usually x~\&~y.
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