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Math Help - Need help here with derivatives

  1. #1
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    Need help here with derivatives

    Hey guys. I'm having trouble with derivatives and I need someone's help with my homework. Could you guys please explain to me how to solve these problems?

    1.
    Given y = f(x) = \frac{1}{x^2} - \sqrt[3]{x}
    Find the second derivative
    f"(x)=\frac{d^2y}{dx^2}

    2.
    Given f(x) = \frac{x^2}{1-3x}
    Find the derivative f'(x)

    3.
    Given \Delta y = 2x^2(\Delta x+3x(\Delta x)^2-2(\Delta x)
    Find the derivative = Lim \frac{f(x+\Delta x)-f(x)}{\Delta x}

    And I'm also stuck at this problem:

    4.
    Given y=f(x)=2x^2-x
    Find \Delta y=f(x+\Delta x)-f(x)

    Help is greatly appreciated.

    Thanks guys!

    --RS
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  2. #2
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    1.) can be written as f(x) = x^(-2) - x^(1/3),
    so the derivative will be:

    f '(x) = -2x^(-3) - (1/3)x^(-2/3)

    2.) you can you the quotient rule, where dy/dx = [V(du/dx) - U(dv/dx)] / V^2 and u = x^2, v = (1-3x)
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  3. #3
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    Quote Originally Posted by Greengoblin View Post
    1.) can be written as f(x) = x^(-2) - x^(1/3),
    so the derivative will be:

    f '(x) = -2x^(-3) - (1/3)x^(-2/3)

    2.) you can you the quotient rule, where dy/dx = [V(du/dx) - U(dv/dx)] / V^2 and u = x^2, v = (1-3x)
    If it is okay by you, could you explain to me how exactly did you do that? I'm sorry, derivatives are one of a math topic I'm stuck with and my professor is those types that knows his stuff but can't really teach it to us.
    Last edited by topsquark; February 14th 2008 at 11:36 AM.
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  4. #4
    Member Greengoblin's Avatar
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    Sure. First of all, the derivative of a sum of functions is the sum of derivatives of those functions, so:

    y = f(x) + g(x) + h(x)
    dy/dx = f '(x) + g '(x) + h '(x)


    The power rule tells us that for a function like x^n:

    dy/dx = nx^(n-1)

    In other words we bring the power down in front of x, and then decrease the power by one.

    when you have the inverse of a function x^n you have 1/ x^n, and this can also be written as x^-n. For the first term I did it like this, and then used the power rule, to get -nx^(-n-1).

    For the second term you can wite square roots as fractions. For example sqrt(x) can be written x^(1/2), the cubed root can be written x^(1/3), and the nth root can be written x^(1/n). Once you have written it like this you can simply apply the power rule to it again. so than if f(x)= nth root of x:

    f(x) = x^(1/n)
    dy/dx = (1/n)x^(1/n - 1)

    When you have a problem like this, it's much easier to write inverses out as negative powers, and square roots out as fractional powers first. Knowing this, you could probably make a good attempt at the second derivative? If not I'll try it for you. Cheers!
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  5. #5
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    Quote Originally Posted by RedSpades View Post
    [snip]
    2.
    Given f(x) = \frac{x^2}{1-3x}
    Find the derivative f'(x)
    [snip]
    The quotient rule says that if y = u(x) v(x), then \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}.

    In your problem, u = x^2 \Rightarrow \frac{du}{dx} = 2x \, and v = 1 - 3x \Rightarrow \frac{dv}{dx} = -3.

    You substitute these four things into the rule, simplifying where necessary. So it's nothing more than basic algebra now.
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