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Thread: 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 $\displaystyle f(t) = t - 1$. 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: $\displaystyle f(x) \times g(x) = (fg)(x)$.

    So $\displaystyle \frac{d}{dt}(f^2(t))$ 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 $\displaystyle [f(x)]^2$.
    Example: $\displaystyle \sin ^2 (x) = \left[ {\sin (x)} \right]^2$
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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 $\displaystyle 2x^2$. That's easy, just use the power rule: $\displaystyle 4x$.

    Now say you want to take the derivative of a function of y where y(x) is a function of x: $\displaystyle 2y^2$. Now we need to include in this the derivative of y in the following manner: $\displaystyle 4y \cdot \frac{dy}{dx}$. 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
    $\displaystyle (f^2(x)+g^3(x))$<-- We'll ignore the x = 2 for now.

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

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

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

    Now apply the x = 2 condition.

    -Dan
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