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Math Help - Using Derivative rules...

  1. #1
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    Using Derivative rules...

    I'm asked to find the slope of the tangent to the graph of the function at the point whose x-coordinate is given...

    for the function: f(x)=x^-3(x^(-1)+1); x=1.

    So simplifying...

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

    Where do I go from here?
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  2. #2
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    Quote Originally Posted by Jeavus View Post
    I'm asked to find the slope of the tangent to the graph of the function at the point whose x-coordinate is given...

    for the function: f(x)=x^-3(x^(-1)+1); x=1.

    So simplifying...

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

    Where do I go from here?
    By f(x)=x^-3(x^(-1)+1) do you mean f(x) = \frac{1}{x^3} \left( \frac{1}{x} + 1 \right) = \frac{1 + x}{x^4}?

    Whether you do or don't, the thing is that you've calculated f(1) when the question requires f'(1), that is, the value of the derivative of f(x) calculated at x = 1.

    If my guess for your function is correct, then my simplification makes it clear that you should apply the quotient rule to get the derivative .....
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