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Math Help - Derivatives

  1. #1
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    Derivatives

    Differentiate the following function:

    \frac{3x}{x^2+4}

    My attempt:
    First I get rid of the fraction:
    f'(x) = (3x)(x^2+4)^-1
    Then I apply product rule F(x) = f(x)g(x) then F'(x) = f(x)g'(x)+f'(x)g(x):
    f'(x) = (3x)(x^2+4)'^-1 + (3x)'(x^2+4)^-1
    I differentiate the derivatives:
    (3x)(2x)^-1 + (3)(x^2+4)^-1
    I simplify:
    \frac{1}{6x^2} + \frac{1}{3x^2+12}

    This is the farthest I can get. What am I doing wrong? Thanks for all the help in advance.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Okay, that is already not the good way around. You haven't found the derivative yet, hence, it's f(x), not f'(x)

    f(x) = (3x)(x^2+4)^{-1}

    Then, since you already used f(x), you should use something else. Let's take u and v, such that:

    f(x) = uv

    f'(x) = uv' + vu'

    u = 3x, v = (x^2 + 4)^{-1}

    u' = 3, v' = -(x^2+4)^{-2}.2x

    Now, substitute that into f'(x).
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  3. #3
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    Quote Originally Posted by Unknown008 View Post
    Okay, that is already not the good way around. You haven't found the derivative yet, hence, it's f(x), not f'(x)

    f(x) = (3x)(x^2+4)^{-1}

    Then, since you already used f(x), you should use something else. Let's take u and v, such that:

    f(x) = uv

    f'(x) = uv' + vu'

    u = 3x, v = (x^2 + 4)^{-1}

    u' = 3, v' = -(x^2+4)^{-2}.2x

    Now, substitute that into f'(x).
    Is this a university level answer? Cause I'm only doing high-school calculus, and have no idea on how I would solve by using your method. Is it impossible to do it the method I was doing? Just when I thought I was getting the hang of it, too.

    Thanks for all the help, by the way. ^^
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  4. #4
    MHF Contributor Unknown008's Avatar
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    No, I'm not in university, though I try to solve some problems from that level, or at least simplify.

    In fact, I'm using your method. Maybe, I should have used f(x) and g(x). I'm just putting f(x), g(x), and their derivatives separately so that you see them more clearly.

    Let y = (3x)(x^2+4)^{-1}

    Then, I apply the rule:

    If y = f(x)g(x)

    Then

    y' = f(x)g'(x) + f'(x)g(x)

    From that, you know that f(x) = 3x and g(x) = (x^2 + 4)^{-1}

    You can find f'(x) and g'(x) form there?

    Then, substitute the values in the rule:

    y' = f(x)g'(x) + f'(x)g(x)

    You already have f(x) and g(x):

    y' = 3xg'(x) + f'(x)(x^2 + 4)^{-1}
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  5. #5
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    There's always the Quotient Rule...

    \displaystyle \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\,\frac{du}{dx} - u\,\frac{dv}{dx}}{v^2}.
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  6. #6
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    Quote Originally Posted by Unknown008 View Post
    No, I'm not in university, though I try to solve some problems from that level, or at least simplify.

    In fact, I'm using your method. Maybe, I should have used f(x) and g(x). I'm just putting f(x), g(x), and their derivatives separately so that you see them more clearly.

    Let y = (3x)(x^2+4)^{-1}

    Then, I apply the rule:

    If y = f(x)g(x)

    Then

    y' = f(x)g'(x) + f'(x)g(x)

    From that, you know that f(x) = 3x and g(x) = (x^2 + 4)^{-1}

    You can find f'(x) and g'(x) form there?

    Then, substitute the values in the rule:

    y' = f(x)g'(x) + f'(x)g(x)

    You already have f(x) and g(x):

    y' = 3xg'(x) + f'(x)(x^2 + 4)^{-1}
    Okay, this is what I get:
    After differentiating g'(x) I get: 2x^{-1} and after differentiating f'(x) I get: 3

    Then I substitute them in:
    F'(x) = (3x)(2x)^{-1} + (3)(x^2+4)^{-1} => \frac{3x}{2x} + \frac{3}{x^2+4} . Do I just cross multiply thereafter and I'm done?

    Gah, rational functions seems to have always been my downfall in mathematics.
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  7. #7
    MHF Contributor Unknown008's Avatar
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    No, g'(x) is wrong.

    For a function of the form:

    y = (f(x))^n

    y' = n(f(x))^{n-1}.f'(x)

    Which means,

    g'(x) = -1.(x^2 + 4)^{-1-1}.2x = -2x(x^2 + 4)^{-2}

    Then, you put it in F'(x) as you did. However, you don't cross multiply. You usually can simplify
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  8. #8
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    Quote Originally Posted by Prove It View Post
    There's always the Quotient Rule...

    \displaystyle \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\,\frac{du}{dx} - u\,\frac{dv}{dx}}{v^2}.
    Unfortunately, my calculus teacher is a really stubborn and stern. He prefers us to only use his method despite the fact that other methods are more widespread and easier for students. He'll give us 0 on tests and assignments for using "foreign methods" as he likes to call it.
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  9. #9
    MHF Contributor Unknown008's Avatar
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    I'm sure that this is the next method you'll learn about in your class, as soon as you're confortable with the product rule, which is what you just used
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  10. #10
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    Quote Originally Posted by Unknown008 View Post
    No, g'(x) is wrong.

    For a function of the form:

    y = (f(x))^n

    y' = n(f(x))^{n-1}.f'(x)

    Which means,

    g'(x) = -1.(x^2 + 4)^{-1-1}.2x = -2x(x^2 + 4)^{-2}

    Then, you put it in F'(x) as you did. However, you don't cross multiply. You usually can simplify
    Okay, thanks for all the help thus far unknown.

    Just one question: How did you get the -2x coefficient in front of (x^2+4)? Doesn't it stay as -1 since you subtracted -1 from -2 after you applied the law?
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  11. #11
    MHF Contributor Unknown008's Avatar
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    The it is -1 multiplied by 2x, giving -2x.

    Now, the 2x is obtained when you differentiate g(x). You first 'lower the power down' next to the bracket, decrease the power by 1 and last but not least, you get the derivative of what is inside the brackets.

    The derivative of x^2 + 4 is 2x.
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  12. #12
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    Okay:

    Now I substitute them in:
    F'(x) = (3x)(2x)^{-1} +(3)(-2(x^2+4))^{-2}
    Then I simplify:
    \frac{3x}{2x} +\frac{3}{(-2x^2-8)^2}
    Now if I were to simplify further, I get the wrong answer seeing as the answer in my book is much different.
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  13. #13
    MHF Contributor Unknown008's Avatar
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    From what you posted earlier, only g'(x) was wrong, hence, correcting to the correct g'(x), we get:

    F'(x) = (3x)(-2x(x^2+4)^{-2}) + (3)(x^2+4)^{-1}

    Then you try simplifying
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  14. #14
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    Thanks.

    This is what I got:
    \frac{3x^3+12x+12x^4+96x^2+192}{(-2x-8)^2(x^2+4)}

    I tried simplifying that and yielded the wrong answer. What have I done wrong?
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  15. #15
    MHF Contributor Unknown008's Avatar
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    Hm... not really, unfortunately.

    F'(x) = (3x)(-2x(x^2+4)^{-2}) + (3)(x^2+4)^{-1}

    F'(x) = \dfrac{(3x)(-2x)}{(x^2+4)^2} + \dfrac{3}{x^2+4}

    There, what you can do is factor the 3/(x^2 + 4)^2:

    F'(x) = \dfrac{3}{(x^2 + 4)^2}\left(-2x^2 +(x^2+4)\right)

    F'(x) = \dfrac{3}{(x^2 + 4)^2}\left(4-x^2\right)

    F'(x) = \dfrac{3(4-x^2)}{(x^2 + 4)^2}

    F'(x) = \dfrac{12-3x^2}{(x^2 + 4)^2}
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