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Math Help - Derivatives problem help.

  1. #1
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    Derivatives problem help.

    How would I start for a problem like this?

    Can anyone solve this problem with steps. I don't know what to do after using the quotient and product rule on the problem below.

    Find the derivative of the function.
    g(x) = (4 + 5x)^5(4 + x - x^2)^6


    Another problem I am stuck on is the one below. How would I start the problem, can anyone show the steps to solve the problem?

    Find the derivative of the function.
    G(y)= (y-2)^4/(y^2+4y)^9


    Thanks for any replies, it would be great to get a reply as soon as possible.
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  2. #2
    MHF Contributor harish21's Avatar
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    after using product rule for the first one, you should get:

    5(4+5x)^4 \times (5) \times (4 + x - x^2)^6 + 6(4 + x - x^2)^5 \times (1-2x) \times (4+5x)^5

     = 25(4+5x)^4(4 + x - x^2)^6+6(1-2x)(4 + x - x^2)^5(4+5x)^5
    Further simplification can be done.


    What problem did you have in the second one? Its a straight application of Quotient Rule plus chain rule... SHow your work if you have any problem..
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  3. #3
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    How would I simplify the first product rule further, would I multiply some things out?

    For the second problem, I get stuck after this step.

    (y^2 + 4y)^9 4(y − 2)^3 1 − (y − 2)^4 9(y^2 + 4y)^8(2y + 4)/
    [(y^2 + 4y)^9]^2
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  4. #4
    MHF Contributor harish21's Avatar
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    = 25(4+5x)^4(4 + x - x^2)^6+6(1-2x)(4 + x - x^2)^5(4+5x)^5

    Take (4+5x)^4 \mbox{and} (4 + x - x^2)^5 as a common factor:

    = (4+5x)^4(4 + x - x^2)^5[25(4 + x - x^2)+6(1-2x)(4+5x)]


    For the second one,

    your notations are hard to understand, but I think you are close to the answer. It is:

     \displaystyle{\frac{[(y^2+4y)^9 \times 4(y-2)^3] - [(y-2)^4 \times 9(y^2+4y)^8 \times(2y+4)]}{(y^2+ 4y)^{18}}}
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  5. #5
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    My book did a sample problem in a different way, how would I get the answer into this form?

    Here are the steps. The way it did the problem confused me.

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  6. #6
    MHF Contributor harish21's Avatar
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     \displaystyle{\frac{[(y^2+4y)^9 \times 4(y-2)^3] - [(y-2)^4 \times 9(y^2+4y)^8 \times(2y+4)]}{(y^2+ 4y)^{18}}}

    In the numerator take  (y^2+4y)^8 \mbox{and} (y-2)^3 as common terms:

     = \displaystyle{\frac{(y^2+4y)^8(y-2)^3[4(y^2+4y)-9(y-2)(2y+4)]}{(y^2+ 4y)^{18}}}

    Rest of the multiplication and cancellation is for you...
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  7. #7
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    Thanks for the help, can you simplify the [25 times etc stuff] in the first problem to get a more simplified answer or is the answer simplified enough?
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  8. #8
    MHF Contributor harish21's Avatar
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    [25(4 + x - x^2)+6(1-2x)(4+5x)]

    You can multiply and add up like terms here..
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  9. #9
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    Would you approach problems like y=arcsin(5x) the same way with the product rule?
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  10. #10
    MHF Contributor harish21's Avatar
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    Product rule????

    If you are talking about differentiation here, you MUST have heard of DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS.
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  11. #11
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    So it's a special way of doing the derivative?

    I've heard of it.
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  12. #12
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by ioke09 View Post
    So it's a special way of doing the derivative?

    I've heard of it.

    What class are you taking? Isnt it calculus I?? Anyways, this link might help
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  13. #13
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    Find y' by implicit differentiation.
    cos(x) + √y = 6

    I'm not sure if I did this correctly, am I on the right track?

    I have -sinx+ 1/2y^-1/2 y'=0 so far.
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