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Math Help - Chain Rule, and Power Rule

  1. #1
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    Chain Rule, and Power Rule

    Can anyone help me out... I know that on problem like  5(5+5x^2)^5  y= 5u^5 , u=5+5x^2 you find the derivtive of those using simple differentiation, and them combine and so on.

    However, I'm not sure how to do this: find the equation of a tangent line to the graph of f using x=-2 on an equation of  f(x)=6x^2(3-6x^2)^5

    I thought that was the derivative, how can this be accomplished. Simply plugging in -2 did not work. .
    Last edited by skyslimit; September 23rd 2008 at 10:38 AM.
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  2. #2
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    I'm confused a bit by your wording. Are you saying f(x)=6x^2..... and you need to find the tangent line to f(x) at x=-2?
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  3. #3
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    Sorry, I should of said: Find an equation for the tangent line to the graph f at x=-2
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  4. #4
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    Ok, so f(x)=6x^2(3-6x^2)^5

    Plugging in x=-2 gives you the y-value of this point on the function. So you'll have a point (-2, f(-2)).

    Now take the derivative of f(x) and plug in -2 to find the slope of the tangent line at x=-2.

    Now you have a point and a slope, thus a line can be formed!

    Point-slope formula: y-y_0=m(x-x_0)

    (x_0 , y_0) is the point on the graph you found and m is the slope at x=-2.

    Make sense?
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  5. #5
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    How can the derivitive of this function be solved using the chain or power rule?
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  6. #6
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    Ok so f(x)=6x^2(3-6x^2)^5. We have a product here, thus the product rule, and the second term is a composite expression thus the chain rule is needed.

    f'(x) = 6x^2 \times \frac{d}{dx} \left( 3-6x^2 \right)^5 + (3-6x^2)^5 \times \frac{d}{dx} \left( 6x^2 \right)

    f'(x)=6x^2[5(3-6x^2)^4 \times (-12x)] + (3-6x^2)^5 \times (12x).
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