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Math Help - Simple Differentiation Question

  1. #1
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    Simple Differentiation Question

    Hi,

    I've hit abit of a roadblock. I was wondering, how would I solve this (using the differntiation rules, not by first principles.) I can't see how to approach this. Could someone show working (such as expanding / factorising) in order to be able to use those rules to solve:


    Differentiate with respect to x: \frac{4x^2+2x}{x^2}


    Thanks!
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  2. #2
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    Quote Originally Posted by Rudey View Post
    Hi,

    I've hit abit of a roadblock. I was wondering, how would I solve this (using the differntiation rules, not by first principles.) I can't see how to approach this. Could someone show working (such as expanding / factorising) in order to be able to use those rules to solve:


    Differentiate with respect to x: \frac{4x^2+2x}{x^2}


    Thanks!
    Note that \frac{4x^2+2x}{x^2}=\frac{1}{x^2}{(4x^2+2x)}.

    So, we can distribute the \frac{1}{x^2} like so:

    =4+\frac{2}{x}=4+2x^{-1}.

    So, can you find \frac{d}{dx}[4+2x^{-1}] ?
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  3. #3
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    Hi. Thanks for reply. I can find the derivative of 4 + 2x^-1 = -2x^x-2

    I'm still confused as what happens with distributing the 1/x^2. I don't see how you go that? Could you explain?

    Sorry about that
    Thanks!
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Rudey View Post
    Hi. Thanks for reply. I can find the derivative of 4 + 2x^-1 = -2x^-2

    I'm still confused as what happens with distributing the 1/x^2. I don't see how you go that? Could you explain?

    Sorry about that
    Thanks!
    Yeah, no problem

    So, take a look at \frac{a+b}{c}.

    It should be clear that this is equivalant to \frac{a}{c}+\frac{b}{c}

    And that this is equivalent to \frac{1}{c}a+\frac{1}{c}b.

    Now, since \frac{1}{c} is common to both terms, I can factor it out giving me

    \frac{1}{c}(a+b).

    Can you see how this applies to your problem? Do you remember this from algebra. It is calle the Distributive Law. It's a biggie. And you're gonna need it.
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  5. #5
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    Hi. Ah yeah I remember that.

    So when differentiating this do you differentiate \frac{1}{x^2} and multiple it by the derivative of 4x^2+2x?
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  6. #6
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    Quote Originally Posted by Rudey View Post
    Hi. Ah yeah I remember that.

    So when differentiating this do you differentiate \frac{1}{x^2} and multiple it by the derivative of 4x^2+2x?
    No.

    \frac{4x^2 + 2x}{x^2} = 4 + \frac{2}{x}.


    So you just differentiate 4 + \frac{2}{x}.


    If you didn't want to simplify at the beginning, you would have to use the Quotient Rule, but that's a lot messier...
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  7. #7
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    Quote Originally Posted by Rudey View Post
    Hi. Ah yeah I remember that.

    So when differentiating this do you differentiate \frac{1}{x^2} and multiple it by the derivative of 4x^2+2x?
    No. All of the algebra was done as to simplify the expression so that taking the derivative would be easy.

    Note that there are a whole bunch of ways to differentiate something:

    The Quotient Rule is one way we coulda went:

    \frac{dy}{dx}=\frac{x^2\cdot(8x+2)-(4x^2+2x)\cdot2x}{(x^2)^2}=\frac{8x^3+2x^2-8x^3-4x^2}{x^4}=-2x^{-2}


    But why go through all that trouble if we coulda simplified the expression first, and the used the power rule, which is so much easier.
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  8. #8
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    Oh. I get it!!! Thank you both so much!!!
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