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Math Help - chain rule problems

  1. #1
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    chain rule problems

    I have been having trouble with the ADVANCED portion of the chain rule and I would like some help on these.
    Find the derivative of these equations
    1. (sinx^2)(cosx^2) The answer comes out to be 2x cos2x^2
    2. (x^3)(2-3x)^2. I know the first is 3x^2 and for the deriv. of the other one, it would be 2(2-3x)(2-3x)'-> (3x^2)(2(2-3x)-3). But the answer is 3x^2(2-3x)(2-5x).
    3.(sqrt x^2+3/x^2-5) The answer comes out to (1/2)(x^2+3/x^2-5)^-1/2(-16x/(x^2-5)^2)
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  2. #2
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    Quote Originally Posted by driver327 View Post
    I have been having trouble with the ADVANCED portion of the chain rule and I would like some help on these.
    Find the derivative of these equations
    1. (sinx^2)(cosx^2) The answer comes out to be 2x cos2x^2)
    \sin x^2 \cos x^2 = \frac{1}{2} \sin (2x^2)
    Derivative,
    \frac{1}{2} \cdot (2x) \cos (2x^2) = x\cos (2x^2)
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  3. #3
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    Quote Originally Posted by driver327 View Post
    2. (x^3)(2-3x)^2. I know the first is 3x^2 and for the deriv. of the other one, it would be 2(2-3x)(2-3x)'-> (3x^2)(2(2-3x)-3). But the answer is 3x^2(2-3x)(2-5x).
    (x^3(2-3x)^2)^{\prime} = (x^3)^{\prime}(2 - 3x)^2 + x^3 ((2 - 3x)^2)^{\prime}

    Now,
    (x^3)^{\prime} = 3x^2
    and
    ((2 - 3x)^2)^{\prime} = 2(2 - 3x) \cdot -3 = -6(2 - 3x)

    So

    (x^3(2-3x)^2)^{\prime} = (x^3)^{\prime}(2 - 3x)^2 + x^3 ((2 - 3x)^2)^{\prime} = 3x^2 \cdot (2 - 3x)^2 + x^3 \cdot -6(2 - 3x)

    You can simplify it from here.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by driver327 View Post
    3.(sqrt x^2+3/x^2-5) The answer comes out to (1/2)(x^2+3/x^2-5)^-1/2(-16x/(x^2-5)^2)
    Please use parenthesis. I presume the problem is to find the derivative of
    \sqrt{\frac{x^2 + 3}{x^2 - 5}}

    So
    \left ( \sqrt{\frac{x^2 + 3}{x^2 - 5}} \right ) ^{\prime} = \left ( \left ( \frac{x^2 + 3}{x^2 - 5} \right ) ^{1/2} \right ) ^{\prime}

    = \frac{1}{2} \cdot \left ( \frac{x^2 + 3}{x^2 - 5} \right ) ^{-1/2} \cdot \left ( \frac{x^2 + 3}{x^2 - 5} \right ) ^{\prime}

    Now,
    \left ( \frac{x^2 + 3}{x^2 - 5} \right ) ^{\prime} = \frac{2x \cdot (x^2 - 5) - (x^2 + 3) \cdot 2x}{(x^2 - 5)^2}

    = \frac{2x^3 - 10x - 2x^3 - 6x}{(x^2 - 5)^2} = \frac{-16x}{(x^2 - 5)^2}

    So finally:
    \left ( \sqrt{\frac{x^2 + 3}{x^2 - 5}} \right ) ^{\prime} = \frac{1}{2} \cdot \left ( \frac{x^2 + 3}{x^2 - 5} \right ) ^{-1/2} \cdot \frac{-16x}{(x^2 - 5)^2}

    Again, I'll let you simplify it from here.

    -Dan
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