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Math Help - Find Derivative of this function

  1. #1
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    Find Derivative of this function

    I need to find a few derivatives for my math homework quickly. Help is appreciated!

    The first one:find f '(x) for the function f(x) = x4 /5 −x9 /5

    And the Second one:

    Find f'(x) for the function c= -870/q +3700 * [e^(2q+3)/820]/q

    Im especially having problems with the second one...
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    Quote Originally Posted by dunbeacrab View Post
    I need to find a few derivatives for my math homework quickly. Help is appreciated!

    The first one:find f '(x) for the function f(x) = x4 /5 −x9 /5

    And the Second one:

    Find f'(x) for the function c= -870/q +3700 * [e^(2q+3)/820]/q

    Im especially having problems with the second one...
    i think you mean:

    1. f(x) = (1/5)x^4 - (1/5)x^9

    and

    2. c = -820/q + [3700*e^{(2q+3)/820}]/q


    1. f(x) = (1/5)x^4 - (1/5)x^9
    => f'(x) = (4/5)x^3 - (9/5)x^8 ............by the power rule

    2. c = -820/q + [3700*e^{(2q+3)/820}]/q

    there are two ways you can do this. change /q to q^-1 and use the power rule with the product rule on the second term. or combine the fractions with q as the common denominator and use the quotient rule. i prefer using the power rule with product rule.

    => c = -820q^-1 + 3700*e^{(2q+3)/820}*q^-1
    => c' = 820q^-2 + 3700[(2/820)e^{(2q+3)/820}*q^-1 - e^{(2q+3)/820}*q^-2]
    => c' = 820q^-2 + (370/41)e^{(2q+3)/820}*q^-1 - 3700*e^{(2q+3)/820}*q^-2
    => c' = (820 - 3700*e^{(2q+3)/820})q^-2 + (370/41)e^{(2q+3)/820}*q^-1
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  3. #3
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    sorry i think i might have mislead you. Ill post the full questions for a better understanding of what im trying to do...

    #1

    For f(x) = x^4 /5 −x^9 /5 find the critical numbers and open intervals where the function is increasing and decreasing.
    Solution: Since f′(x)= ___________, we see that f′(x)=0 if x= _______ and that f′( ______ ) does not exist.
    Therefore, the critical numbers are ______ and ______ .





    #2



    .
    C= -870/q + 3700 * [e^(2q+3)/820]/q




    Find the marginal cost if q=97.



    Answer: The marginal cost is _____ dollars/(one unit of the product).


    .

    Now a third one...
    #3
    a. Find y', for y=(8x+7)^7x+1
    b. take natural log of both sides of y'
    c. simplify using properties of logs ( In your answer, lna must be entered as log(a).)
    d. Differentiate both sides with respect to x:
    y′/y = ________
    e.Solve for y′: y′ = y( _______________ ).
    f.Substitute the original expression for y to get y′ in terms of x only:
    y′ = __________________________________


    .
    .

    Sorry for so many now... i just got a deadline and I dont have time to figure all of this out.

    Thanks in advance!
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    Quote Originally Posted by dunbeacrab View Post
    sorry i think i might have mislead you. Ill post the full questions for a better understanding of what im trying to do...

    #1

    For f(x) = x^4 /5 −x^9 /5 find the critical numbers and open intervals where the function is increasing and decreasing.
    Solution: Since f′(x)= ___________, we see that f′(x)=0 if x= _______ and that f′( ______ ) does not exist.
    Therefore, the critical numbers are ______ and ______ .
    wait, according to the derivative we found, f'(x) exists everywhere. is the function actually x^(4/5) - x^(9/5) ?
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  5. #5
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    x^(4/5) - x^(9/5)

    yeah sorry this is the correct function
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    Quote Originally Posted by dunbeacrab View Post
    sorry i think i might have mislead you. Ill post the full questions for a better understanding of what im trying to do...

    #1

    For f(x) = x^(4/5) −x^(9/5) find the critical numbers and open intervals where the function is increasing and decreasing.
    Solution: Since f′(x)= ___________, we see that f′(x)=0 if x= _______ and that f′( ______ ) does not exist.
    Therefore, the critical numbers are ______ and ______ .
    fine. let's start from scratch.

    f(x) = x^(4/5) - x^(9/5)
    => f'(x) = (4/5)x^(-1/5) - (9/5)x^(4/5)........notice this function does not exist for x <= 0. since (4/5)x^(-1/5) = 4/5x^(1/5), x cant be 0. and since (9/5)x^(4/5) has an even root for x, x can't be negative. so f'(x) does not exist on (-infinity, 0]

    now we set f'(x)=0
    => (4/5)x^(-1/5) - (9/5)x^(4/5) = 0
    => (20 - 45x)/(5x^(1/5)) = 0 ...................i combined the fractions
    => 20 - 45x = 0 ....................................since the denominator can't be zero we forget about it and equate the top to zero

    => x = 4/9
    so f'(x) = 0 if x = 4/9

    i suppose the critical numbers are the x and y value of the point.
    in that case they are x=4/9 and y=0.2904
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  7. #7
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    hmm thats not one of the options i get to choose from...the derivative i mean.. any suggestion on the other question?
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    Quote Originally Posted by dunbeacrab View Post
    hmm thats not one of the options i get to choose from...the derivative i mean.. any suggestion on the other question?
    what are the options?
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    Quote Originally Posted by dunbeacrab View Post

    #3
    a. Find y', for y=(8x+7)^7x+1
    b. take natural log of both sides of y'
    c. simplify using properties of logs ( In your answer, lna must be entered as log(a).)
    d. Differentiate both sides with respect to x:
    y′/y = ________
    e.Solve for y′: y′ = y( _______________ ).
    f.Substitute the original expression for y to get y′ in terms of x only:
    y′ = __________________________________
    you have to be clear on what you're typing. i'm not sure if 7x+1 is the power, or just 7x and the 1 is a separate term
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  10. #10
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    1/9*(9-5*x)*x^(-1/9)

    (9-4*x)*x^(-1/9)

    1/5*(4-9*x)*x^(-1/5)

    1/4*(5-9*x)*x^(-1/5)


    one of these....
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  11. #11
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    also how do you differentiate both sides of this equation with respect to X:

    lny=(7x+1)ln(8x+7)

    im confused by this...
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    Quote Originally Posted by dunbeacrab View Post
    also how do you differentiate both sides of this equation with respect to X:

    lny=(7x+1)ln(8x+7)

    im confused by this...
    judging from how the question is phrased i believe you do it by implicit differentiation.

    lny=(7x+1)ln(8x+7)

    => y'/y = 8(7x + 1)/(8x + 7) + 7ln(8x + 7)
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    Quote Originally Posted by dunbeacrab View Post
    also how do you differentiate both sides of this equation with respect to X:

    lny=(7x+1)ln(8x+7)

    im confused by this...
    sorry, i'm in a rush so i cant really explain things very well right now.

    you find the derivative of both sides. whenever you find the derivative of y you attach a y' to it.

    so derivative of lny = 1/y y' = y'/y ...........it's like the y' multiplies. the other side you differentiate normally by the product rule
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  14. #14
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    err... tell me where im going wrong...

    y'/y= u(x)*v'(x)+v(x)*u'(x)

    = (7x+1)*(1/8x+7) + (8x+7)*7

    = (7x+1)/(8x+7) + 7*(8x+7)


    Where have i gone wrong?
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    Quote Originally Posted by dunbeacrab View Post
    err... tell me where im going wrong...

    y'/y= u(x)*v'(x)+v(x)*u'(x)

    = (7x+1)*(1/8x+7) + (8x+7)*7

    = (7x+1)/(8x+7) + 7*(8x+7)


    Where have i gone wrong?
    you left off the ln for the second piece, and you forgot to multiply the first piece by the derivative of 8x + 7 to complete the chain rule. i did this question earlier
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