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Math Help - Derivative of a function!

  1. #1
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    Derivative of a function!

    How would you find the derivative for these three problems?? I end up with different answers each time I do them.

    y= (4x^2+7)^2 * (3x^3+1)^4

    y= (2x)^(1/2) + (2/x)^(1/2)

    y= (x^2-5)^(1/2) * (x^2+3)^(1/3)

    I'd be happy if anyone could help! : )
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by coolaid2u View Post
    How would you find the derivative for these three problems?? I end up with different answers each time I do them.

    y= (4x^2+7)^2 * (3x^3+1)^4

    y= (2x)^(1/2) + (2/x)^(1/2)

    y= (x^2-5)^(1/2) * (x^2+3)^(1/3)

    I'd be happy if anyone could help! : )
    y= (4x^2+7)^2 * (3x^3+1)^4 ............we need the chain rule and product rule here
    => y' = 2(4x^2 + 7)(8x)(3x^3 + 1)^4 + (4x^2+7)^2 * 4(3x^3 + 1)^3 * 9x^2
    => y' = 16x(4x^2 + 7)(3x^3 + 1)^4 + (36x^2)(4x^2+7)^2 * (3x^3 + 1)^3
    => y' = [4x(4x^2 + 7)(3x^3 + 1)^3]* [4(3x^3 + 1) + 9x(4x^2+7)]



    y= (2x)^(1/2) + (2/x)^(1/2) .......we need the chain rule for each of these
    => y' = (1/2)(2x)^(-1/2) * 2 + (1/2)(2/x)^(-1/2) * (-2x^-2)
    => y' = (2x)^(-1/2) - (x^-2)(2/x)^(-1/2)



    y= (x^2-5)^(1/2) * (x^2+3)^(1/3)
    .......product rule combined with chain rule again

    => y' = (1/2)(x^2 - 5)^(-1/2) * (2x)(x^2+3)^(1/3) + (x^2-5)^(1/2) * (1/3)(x^2 + 3)^(-2/3) * (2x)
    => y' = x(x^2 - 5)^(-1/2) * (x^2+3)^(1/3) + (2/3)x(x^2-5)^(1/2) * (x^2 + 3)^(-2/3)
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    Thank You!!!!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by coolaid2u View Post
    Thank You!!!!
    did you get the steps? do you know how to apply the product and chain rules?
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  5. #5
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    Yes, I just wrote the rules wrong . Thanks!
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