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Math Help - derivative evaluate

  1. #1
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    derivative evaluate

    Im confused on what this is telling me.
    d/dt (f^2(t)) at t=2 i know I have to find the derivative but how with the f^2 Ive never seen that before could someone please explain.
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  2. #2
    TD!
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    It looks like there's a given function f, and you are asked to find the derivative of its square, f, with respect to t in t = 2.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ocmisssunshine View Post
    Im confused on what this is telling me.
    d/dt (f^2(t)) at t=2 i know I have to find the derivative but how with the f^2 Ive never seen that before could someone please explain.
    Say, for example f(t) = t - 1. Then the function [tex]f^2(t) = (t - 1) \times (t - 1) = (t - 1)^2[tex].

    This is an extension of the notation for multiplying two functions: f(x) \times g(x) = (fg)(x).

    So \frac{d}{dt}(f^2(t)) is done via the chain rule. Does that suffice? (Hint: the answer will be in terms of f and its derivative.)

    -Dan
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    Its a graph but I cant figure out how to find it form just the graph.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ocmisssunshine View Post
    Its a graph but I cant figure out how to find it form just the graph.
    You are saying you have the graph of y = f(t)?

    You may need to post it for us to help you better.

    -Dan
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    I will try that one on my own cause I cant upload the graph but can you help me with this instead.... this is given...
    f(2)= 4
    f '(2)= 4
    f ''(2)=-1
    g(2)=2
    g '(2)=5

    I have to find the derivative of (f^2(x)+g^3(x)) at x=2 I cant understand the chain rule for the life of me.
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    Quote Originally Posted by ocmisssunshine View Post
    Its a graph but I cant figure out how to find it form just the graph.
    With most graphing utilities one needs to enter [f(x)]^2.
    Example: \sin ^2 (x) = \left[ {\sin (x)} \right]^2
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ocmisssunshine View Post
    I will try that one on my own cause I cant upload the graph but can you help me with this instead.... this is given...
    f(2)= 4
    f '(2)= 4
    f ''(2)=-1
    g(2)=2
    g '(2)=5

    I have to find the derivative of (f^2(x)+g^3(x)) at x=2 I cant understand the chain rule for the life of me.
    The chain rule is a way to take the derivative of a function of a function.

    Say, for example, you want to take the derivative of 2x^2. That's easy, just use the power rule: 4x.

    Now say you want to take the derivative of a function of y where y(x) is a function of x: 2y^2. Now we need to include in this the derivative of y in the following manner: 4y \cdot \frac{dy}{dx}. The first factor is just like the x case; all we've added is the derivative on the end.

    So let's take a look at your problem:
    Take the derivative of
    (f^2(x)+g^3(x))<-- We'll ignore the x = 2 for now.

    Take the derivative of each term separately:
    f^2(x) has a derivative of 2f(x) \cdot \frac{df}{dx}.
    (Note the similarities with the previous example with the y in it.)

    g^3(x) has a derivative of 3g^2(x) \cdot \frac{dg}{dx}.

    So the derivative of the whole expression is
    2f(x) \cdot \frac{df}{dx} + 3g^2(x) \cdot \frac{dg}{dx}

    Now apply the x = 2 condition.

    -Dan
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