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Math Help - please help me

  1. #1
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    please help me

    Hi, I really hate to ask for help as I'm usually good at math, but I have homework due at 10 o'clock tonight and I'm struggling. Could someone please help me with the following problems? I'm not asking for you to do my homework for me as that wouldn't help me get a good grade in the class, I just have no idea how to do these.

    3. The next one is REALLY confusing.

    F(x)=f(x^3) and G(x)=(f(x))^3

    a^2=13
    f(a)=3
    f'(a)=11
    f'(a^3)=15

    I found F'(a), but I can't figure out G'(a)

    I really hate to ask so many questions, but I've been working on this for an hour and I've gotten nowhere on these. Help on ANY of them would be great.

    Thank you.
    Last edited by nobodygirl; November 8th 2006 at 06:42 PM.
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  2. #2
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    Quote Originally Posted by nobodygirl View Post
    Hi, I really hate to ask for help as I'm usually good at math, but I have homework due at 10 o'clock tonight and I'm struggling. Could someone please help me with the following problems? I'm not asking for you to do my homework for me as that wouldn't help me get a good grade in the class, I just have no idea how to do these.

    1. lnx^5 and I need to find the first and second derivatives
    Let f(x) = [ln(x)]^5; I am assuming you mean this? Or, do you mean ln(x^5)?

    I'll do both; I'll assume for case 1 you mean the first and for case 2 you mean the latter:

    Case 1:

    f(x) = [ln(x)]^5
    f'(x) = 5*[ln(x)]^4*(1/x) = [5*[ln(x)]^4]/x
    f''(x) = Use quotient rule:

    Quotient rule states:

    [f(x)/g(x)]' = [g(x)*f'(x) - f(x)*g'(x)]/[g(x)]^2

    [x*20*[ln(x)]^3*(1/x) - [5*[ln(x)]^4]*1]/[x]^2

    = [20*[ln(x)]^3 - 5*[ln(x)]^4]/x^2

    Case 2:

    f(x) = ln(x^5)
    f'(x) = 5/x
    f''(x) = -5/(x^2)

    I think the above is pretty easy to see, so I think you mean the first one.
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  3. #3
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    Quote Originally Posted by nobodygirl View Post
    Hi, I really hate to ask for help as I'm usually good at math, but I have homework due at 10 o'clock tonight and I'm struggling. Could someone please help me with the following problems? I'm not asking for you to do my homework for me as that wouldn't help me get a good grade in the class, I just have no idea how to do these.

    5. find f'(x) when f(x)=-6cos(sin(x^6))
    This is actually easier than it looks; we have a composition of functions.

    f(x) = -6cos(sin(x^6))

    f'(x) = -6*-sin(sin(x^6))*cos(x^6)*6*x^5

    Simplify:

    Thus, f'(x) = 36*x^5*cos(x^6)*sin(sin(x^6))
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  4. #4
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    I was talking about the first way.

    I tried putting in the answer you had for that one, and it said that it was incorrect. My question is how did you get the 5ln^4 times 1/x?
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  5. #5
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    Quote Originally Posted by AfterShock View Post
    This is actually easier than it looks; we have a composition of functions.

    f(x) = -6cos(sin(x^6))

    f'(x) = -6*-sin(sin(x^6))*cos(x^6)*6*x^5

    Simplify:

    Thus, f'(x) = 36*x^5*cos(x^6)*sin(sin(x^6))
    Ah, thanks, I just made several stupid mistakes in that one.
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  6. #6
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    Quote Originally Posted by nobodygirl View Post
    Hi, I really hate to ask for help as I'm usually good at math, but I have homework due at 10 o'clock tonight and I'm struggling. Could someone please help me with the following problems? I'm not asking for you to do my homework for me as that wouldn't help me get a good grade in the class, I just have no idea how to do these.

    4. I need to find the f'(x) when f(x)=8e^(xcosx)
    f(x) = 8e^(xcosx)

    Another easy one;

    f'(x) = 8*e^(x*cos(x))*{derivative of inside of composition here = g(x)}

    g(x) = x*cos(x)
    g'(x) = 1*cos(x) - x*sin(x)

    Thus, f'(x) = 8*e^(x*cos(x))*(cos(x) - x*sin(x))
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  7. #7
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    Quote Originally Posted by AfterShock View Post
    f(x) = 8e^(xcosx)

    Another easy one;

    f'(x) = 8*e^(x*cos(x))*{derivative of inside of composition here = g(x)}

    g(x) = x*cos(x)
    g'(x) = 1*cos(x) - x*sin(x)

    Thus, f'(x) = 8*e^(x*cos(x))*(cos(x) - x*sin(x))
    I ALWAYS forget product rule, thank you so much, you're a life saver!
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  8. #8
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    Quote Originally Posted by nobodygirl View Post
    I was talking about the first way.

    I tried putting in the answer you had for that one, and it said that it was incorrect. My question is how did you get the 5ln^4 times 1/x?
    If you integrate both of them, you will see they are correct. That is, you will end up with f'(x) integrating f''(x) and f(x) integrating f'(x).

    To answer your question:

    f(x) = [ln(x)]^5
    f'(x) = 5*[ln(x)]^4*(1/x) = [5*[ln(x)]^4]/x

    (1/x) is taking the derivative of the inside function; do you see why?

    5*[ln(x)]^4*{derivative of composition here}, in which case we need to find the derivative of ln(x), which is 1/x, thus,

    f'(x) = 5*[ln(x)]^4*(1/x) = [5*[ln(x)]^4]/x
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  9. #9
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    Quote Originally Posted by AfterShock View Post
    If you integrate both of them, you will see they are correct. That is, you will end up with f'(x) integrating f''(x) and f(x) integrating f'(x).

    To answer your question:

    f(x) = [ln(x)]^5
    f'(x) = 5*[ln(x)]^4*(1/x) = [5*[ln(x)]^4]/x

    (1/x) is taking the derivative of the inside function; do you see why?

    5*[ln(x)]^4*{derivative of composition here}, in which case we need to find the derivative of ln(x), which is 1/x, thus,

    f'(x) = 5*[ln(x)]^4*(1/x) = [5*[ln(x)]^4]/x
    I can't believe I didn't realize that I had to actually finish the chain rule.

    You are amazing.
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