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Math Help - differentiate!

  1. #1
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    differentiate!

    1) f(x)=(7x+\sqrt{x^2+6})^4

    2) f(x)=tan^3 (4x)

    3) g(x)=(\frac{x-2}{2x+2})^9

    4) p(x)= [3x(5x-1)^5]^4

    I tried to work out all 4 of these but I keep getting the wrong answer.
    Last edited by yoman360; September 1st 2009 at 04:04 PM.
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  2. #2
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    Quote Originally Posted by yoman360 View Post
    1) f(x)=(7x+\sqrt{x^2+6})^4

    2) f(x)=tan^3 (4x)

    3) g(x)=(\frac{x-2}{2x+2})^9

    4) p(x)= [3x(5x-1)^5]^4

    I tried to work out all 4 of these but I keep getting the wrong answer.
    all involve the chain rule ...

    1) f'(x) = 4(7x + \sqrt{x^2+6})^3 \cdot \left(7 + \frac{x}{\sqrt{x^2+6}}\right)


    2) f'(x) = 3\tan^2(4x) \cdot \sec^2(4x) \cdot 4


    3) g'(x) = 9\left(\frac{x-2}{2x+2}\right)^8 \cdot \frac{(2x+2)(1) - (x-2)(2)}{(2x+2)^2}


    4) p'(x)= 4[3x(5x-1)^5]^3 \cdot [3x \cdot 5(5x-1)^4 \cdot 5 + (5x-1)^5 \cdot 3]


    you can clean up the algebra.
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  3. #3
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    Quote Originally Posted by skeeter View Post
    all involve the chain rule ...

    1) f'(x) = 4(7x + \sqrt{x^2+6})^3 \cdot \left(7 + \frac{x}{\sqrt{x^2+6}}\right)


    2) f'(x) = 3\tan^2(4x) \cdot \sec^2(4x) \cdot 4


    3) g'(x) = 9\left(\frac{x-2}{2x+2}\right)^8 \cdot \frac{(2x+2)(1) - (x-2)(2)}{(2x+2)^2}


    4) p'(x)= 4[3x(5x-1)^5]^3 \cdot [3x \cdot 5(5x-1)^4 \cdot 5 + (5x-1)^5 \cdot 3]


    you can clean up the algebra.
    Thanks I kinda don't understand the chain rule but as I do more problems it might make more sense.
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