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Math Help - Calculus HW due in 3 hours (4 Problems I cant get)

  1. #1
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    Calculus HW due in 3 hours (4 Problems I cant get)

    Well I've been trying these problems for awhile, I'll type the problem out as well as what I have gotten as the answer and how I got there. Thanks for any help...my prof likes to throw things at us that is very challengin on HW..
    PROBLEM 1
    f(x)=1/(x-2x^2)^(9/8)
    Find the derivative of f(x).

    For starters i moved the binomial to the numerator and switched the power to negative giving me f(x)=(x-2x^2)^(-9/8) . After that I simply applied the chain rule giving me (-9/8)(x-2x^2)^-17/8(-4x) . It says this answer is incorrect...

    PROBLEM 2
    f(x)=sqrt(3+sqrt4x) The 4x has two sqrt over it. I broke this problem down to read f(x)=(3^1/2+4x^1/2)^1/2 . I then used the chain rule which gave me (1/2)(3^1/2+4x^1/2)(2x^-1/2) . It says this answer is incorrect...

    I'll post the other two up after I get some help on these because I havent even been able to start them. They are more difficult then these two. Thanks for the help.

    -Brad
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  2. #2
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    Hello, Brad!

    You made some small errors . . .


    PROBLEM 1: .f(x) .= .1/(x - 2x2)^{9/8}
    Find the derivative of f(x).

    For starters i moved the binomial to the numerator and switched the power
    to negative, giving me: .f(x) .= .(x - 2x)^{-9/8}

    After that I simply applied the chain rule giving me: .(-9/8)(x - 2x)^{-17/8}(-4x) . no
    The derivative of the "inside" is: .1 - 4x


    PROBLEM 2: .f(x) .= .sqrt[3 + sqrt(4x)]

    I broke this problem down to: .f(x) .= .(3 + 4x^)^ . no
    The inner square root is on the entire 4x.

    . . f(x) .= .[3 + (4x)^]^ .= .(3 + 2x^)^

    Now give it a try . . .

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  3. #3
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    Quote Originally Posted by lmao View Post
    Well I've been trying these problems for awhile, I'll type the problem out as well as what I have gotten as the answer and how I got there. Thanks for any help...my prof likes to throw things at us that is very challengin on HW..
    PROBLEM 1
    f(x)=1/(x-2x^2)^(9/8)
    Find the derivative of f(x).

    For starters i moved the binomial to the numerator and switched the power to negative giving me f(x)=(x-2x^2)^(-9/8) . After that I simply applied the chain rule giving me (-9/8)(x-2x^2)^-17/8(-4x) . It says this answer is incorrect...

    PROBLEM 2
    f(x)=sqrt(3+sqrt4x) The 4x has two sqrt over it. I broke this problem down to read f(x)=(3^1/2+4x^1/2)^1/2 . I then used the chain rule which gave me (1/2)(3^1/2+4x^1/2)(2x^-1/2) . It says this answer is incorrect...

    I'll post the other two up after I get some help on these because I havent even been able to start them. They are more difficult then these two. Thanks for the help.

    -Brad
    Q1 f(x)=(x-2x^2)^(-9/8)

    Let u = (x-2x^2)
    u' = 1-4x
    f(x) = u^(-9/8)
    df(x)/du = (-9/8)u^(-17/8)

    Then df(x)/dx = (1-4x)(-9/8)u^(-17/8)
    = (1-4x)(-9/8)[(x-2x^2)^(-17/8)]

    Q2 f(x)=sqrt(3+sqrt4x) = [ 3+((4x)^1/2) ] ^ (1/2)

    Let u = 3+((4x)^1/2)
    u' = (2x)^(-1/2)
    f(x) = u ^ (1/2)
    df(x)/du = (1/2)u^(-1/2)

    Then df(x)/dx = (2x)^(-1/2) . (1/2)u(-1/2)
    = (2x)^(-1/2) . (1/2)[3+((4x)^1/2)]^(-1/2)

    That's very ugly, but I'm sure you can simplify it in some way.
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  4. #4
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    Quote Originally Posted by lmao View Post
    Well I've been trying these problems for awhile, I'll type the problem out as well as what I have gotten as the answer and how I got there. Thanks for any help...my prof likes to throw things at us that is very challengin on HW..
    PROBLEM 1
    f(x)=1/(x-2x^2)^(9/8)
    Find the derivative of f(x).

    For starters i moved the binomial to the numerator and switched the power to negative giving me f(x)=(x-2x^2)^(-9/8) . After that I simply applied the chain rule giving me (-9/8)(x-2x^2)^-17/8(-4x) . It says this answer is incorrect...

    PROBLEM 2
    f(x)=sqrt(3+sqrt4x) The 4x has two sqrt over it. I broke this problem down to read f(x)=(3^1/2+4x^1/2)^1/2 . I then used the chain rule which gave me (1/2)(3^1/2+4x^1/2)(2x^-1/2) . It says this answer is incorrect...

    I'll post the other two up after I get some help on these because I havent even been able to start them. They are more difficult then these two. Thanks for the help.

    -Brad
    For Problem 1.
    The derivative of (x -2x^2) is (1 -4x). Not (-4x) only.

    For Problem #2.
    Your f(x) = [(3^(1/2) +4x^(1/2)]^(1/2) is really incorrect.
    What is the name of that property? Sorry, I am not well-versed on definitions, but you cannot "spread" the sqrt to each term in the radicand that are in summation.
    sqrt(9 +16) is not sqrt(9) + sqrt(16). -----that gives 3 +4 = 7.
    sqrt(9 +16) is sqrt(25). -----that gives 5, which is correct.

    f(x) = sqrt[3 +sqrt(4x)]
    So,
    f(x) = [3 +(4x)^(1/2)]^(1/2)
    then,
    f'(x) = (1/2)[3 +(4x)^(1/2)]^(-1/2) *(1/2)[(4x)^(-1/2) *4]
    f'(x) = (1/2)(1/2)(4)[3 +(4x)^(1/2)]^(-1/2) *[(4x)^(-1/2)]
    f'(x) = [3 +(4x)^(1/2)]^(-1/2) *[(4x)^(-1/2)]
    Since sqrt(a) *sqrt(b) = sqrt(a*b), then,
    f'(x) = [{3 +(4x)^(1/2)}*(4x)]^(-1/2)

    I intentionally did not reduce the sqrt(4x) in 2sqrt(x) yet so that you won't be more confused.
    Here, sqrt(4x) = sqrt(4) *sqrt(x). You can "spread" here the sqrt to each term in the radicand because those terms are in multiplication.
    sqrt(9 +16) is not the same as sqrt(9*16)
    sqrt(9*16) = sqrt(9) *sqrt(16) = 3*4 = 12.
    sqrt(9*16) = sqrt(144) = 12.

    So,
    f'(x) = [{3 +2sqrt(x)}*(2sqrt(x))]^(-1/2)
    f'(x) = [3*2sqrt(x) +[2sqrt(x)]*2sqrt(x)]^(-1/2)
    f'(x) = [6sqrt(x) +4x]^(-1/2)
    f'(x) = 1 / sqrt[6sqrt(x) +4x] ---------------answer.
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  5. #5
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    Quote Originally Posted by lmao View Post
    ...

    PROBLEM 2
    f(x)=sqrt(3+sqrt4x) The 4x has two sqrt over it. I broke this problem down to read f(x)=(3^1/2+4x^1/2)^1/2 . I then used the chain rule which gave me (1/2)(3^1/2+4x^1/2)(2x^-1/2) . It says this answer is incorrect...
    ...
    Hallo, Brad,

    I used the result Soroban has given. Then use the chain rule very carefully:
    Attached Thumbnails Attached Thumbnails Calculus HW due in 3 hours (4 Problems I cant get)-imao_abltg.gif  
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  6. #6
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    Quote Originally Posted by lmao View Post
    Well I've been trying these problems for awhile, I'll type the problem out as well as what I have gotten as the answer and how I got there. Thanks for any help...my prof likes to throw things at us that is very challengin on HW..
    PROBLEM 1
    f(x)=1/(x-2x^2)^(9/8)
    Find the derivative of f(x).

    For starters i moved the binomial to the numerator and switched the power to negative giving me f(x)=(x-2x^2)^(-9/8) . After that I simply applied the chain rule giving me (-9/8)(x-2x^2)^-17/8(-4x) . It says this answer is incorrect...

    PROBLEM 2
    f(x)=sqrt(3+sqrt4x) The 4x has two sqrt over it. I broke this problem down to read f(x)=(3^1/2+4x^1/2)^1/2 . I then used the chain rule which gave me (1/2)(3^1/2+4x^1/2)(2x^-1/2) . It says this answer is incorrect...

    I'll post the other two up after I get some help on these because I havent even been able to start them. They are more difficult then these two. Thanks for the help.

    -Brad
    You already posted this question in the Calculus section... by double posting, you waste a lot of the helpers' time.
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