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Math Help - Derivative

  1. #1
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    Derivative

    y=|x|(tanx)^(1/2)

    find y'
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  2. #2
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    Quote Originally Posted by Noxide View Post
    y=|x|(tanx)^(1/2)

    find y'
    Are you given a specific values for x?
    or an interval?
    Last edited by General; February 10th 2010 at 09:22 PM.
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  3. #3
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    Quote Originally Posted by Noxide View Post
    y=|x|(tanx)^(1/2)

    find y'
    Run the online derivative calculator:

    webMathematica Explorations: Step-by-Step Derivatives

    Derivative of: x*Tan[x]^(1/2)
    With respect to: x

    Gives:

    \sqrt{\tan(x)} + \frac{x \sec^2(x)}{2\sqrt{\tan(x)}}

    This is valid for x > 0

    For x < 0, you have

    Derivative of: -x*Tan[x]^(1/2)
    With respect to: x

    which gives:

    \sqrt{\tan(x)} - \frac{x \sec^2(x)}{2\sqrt{\tan(x)}}

    So we can combine these two into one expression for x \ne 0:

    \sqrt{\tan(x)} + \left|\frac{x \sec^2(x)}{2\sqrt{\tan(x)}}\right|
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  4. #4
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    thanks, i was on the right track
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  5. #5
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    is it correct to say (with respect to my first question):


    if x = 0 then y = 0 and d0/dx = 0?
    or should i differentiate first using the positive case of x then input the value of x into the derivative and get 0?
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  6. #6
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    Quote Originally Posted by Noxide View Post
    is it correct to say (with respect to my first question):


    if x = 0 then y = 0 and d0/dx = 0?
    or should i differentiate first using the positive case of x then input the value of x into the derivative and get 0?
    Well, if y is a function of x, in other words, y = f(x), you can't do that directly. In this case,

    y(x)=|x|\sqrt{\tan x}

    So we compute the derivative as before to find y'(x) = \frac{dy}{dx}(x). Only once you've done that can you put in x = 0 to find what the value of y'(0) is.

    But because you have an absolute value, at x = 0 the derivative doesn't exist. That's because the slope of the tangent line of |x| changes sign at x=0.
    Last edited by lgstarn; February 12th 2010 at 07:50 AM. Reason: Editing sqrt
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  7. #7
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    Thanks.

    I can picture that there would be an infinite number of tangent lines at that point. So there are infinite derivatives. Is there a way to get a general equation for those infinite derivatives?
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  8. #8
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    Quote Originally Posted by Noxide View Post
    Thanks.

    I can picture that there would be an infinite number of tangent lines at that point. So there are infinite derivatives. Is there a way to get a general equation for those infinite derivatives?
    We just say it isn't differentiable at x=0 and exclude the point x=0 from the domain of the derivative. You don't need to bother doing anything else.

    The domains of the function and its derivative are a bit complicated though (since \sqrt{\tan x} only exists where \tan x \ge 0). Are you asked to state the domain of the derivative?
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  9. #9
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    Quote Originally Posted by Noxide View Post
    Thanks.

    I can picture that there would be an infinite number of tangent lines at that point. So there are infinite derivatives. Is there a way to get a general equation for those infinite derivatives?
    Actually, you are correct! That is the concept of a "weak derivative." This is a bit advanced though. You wouldn't have to see that stuff until you became a graduate student or a senior working in mathematics.

    Don't let the following links blow your mind - just be aware that what you are talking about does exist.

    Weak derivative - Wikipedia, the free encyclopedia
    Subderivative - Wikipedia, the free encyclopedia
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