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Math Help - Chain Rule

  1. #1
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    Post Chain Rule

    Four questions in the Homework are bothering me :


    see attachment cause I had to use word to put the equation
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  2. #2
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    For the most part, it's just differentiation. Find the derivative and enter in the given values.
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    Yea this is what I tried to do except i kept getting the wrong answer
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    Quote Originally Posted by aaasssaaa View Post
    Four questions in the Homework are bothering me :


    see attachment cause I had to use word to put the equation
    let g(x) = 27/(2x+1)^3 Determine the slope of the tangent to the curve at x=1

    g(x) = \frac{27}{(2x+1)^3}

    g'(x) = \frac{-3 \cdot 27}{(2x+1)^4}

    g'(1) = \frac{-3 \cdot 27}{(2 \cdot 1 +1)^4} =  -\frac{81}{81} = -1

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by aaasssaaa View Post
    Four questions in the Homework are bothering me :


    see attachment cause I had to use word to put the equation
    Let f(x) = \frac{6}{5 \sqrt{3x^4} - x}. Find f'(1).

    f(x) = \frac{6}{5 \sqrt{3x^4} - x}

    f'(x) = \frac{-6}{(5 \sqrt{3x^4} - x)^{2}} \cdot \left ( 5 \cdot \frac{1}{2\sqrt{3x^4}} \cdot 3 \cdot 4x^3 - 1 \right )

    f'(x) = \frac{-6 \cdot (10\sqrt{3}x - 1)}{(5\sqrt{3}x^2 - x)^2}

    f'(1) = \frac{-6 \cdot (10\sqrt{3} \cdot 1 - 1)}{(5\sqrt{3}(1)^2 - 1)^2} \approx -1.66878

    -Dan
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    Quote Originally Posted by aaasssaaa View Post
    Four questions in the Homework are bothering me :


    see attachment cause I had to use word to put the equation
    s = 5 - 3t - \left [ t^{-2} + (2t - 5)^5 \right ] ^4

    s' = -3 - 4 \left [ t^{-2} + (2t - 5)^5 \right ] ^3 \cdot \left ( -2t^{-3} + 5(2t - 5)^4 \cdot 2 \right )

    s' = -3 - 4 \left ( -2t^{-3} + 10(2t - 5)^4 \right ) \left [ t^{-2} + (2t - 5)^5 \right ] ^3

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by aaasssaaa View Post
    Four questions in the Homework are bothering me :


    see attachment cause I had to use word to put the equation
    V(t) = 50000 \left ( 1 - \frac{t}{30} \right ) ^2

    V'(t) = 50000 \cdot 2 \left ( 1 - \frac{t}{30} \right ) \cdot \frac{-1}{30}

    V'(t) = - \frac{10000}{3} \left ( 1 - \frac{t}{30} \right )

    V'(10) = - \frac{10000}{3} \left ( 1 - \frac{10}{30} \right )

    V'(10) = - \frac{20000}{9}

    Technically they want 20000/9 gal/min (ie not negative) because the question asked for the rate the water is flowing out.

    -Dan
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