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  1. #1
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    Help and compare answers

    Differentiate the following
    1. (i) \frac{3}{x^2}
    I struggled with these because I'm used to differentiating square roots and fractions.

    (ii) y = x+\sqrt{x}

    2. Find the gradient of the curve y = x^2 + \frac{1}{x} at the point (1,2)
    Answer I got = 3

    3. Differentiate
    (i) y=(x+2)(x^3+5x)
    Answer I got = 5x^3 + 10x - 6x^2 +10

    (ii) y=\frac{3x+2}{2x+1}
    Answer I got = \frac{7}{(2x+1)^2}

    Are my answers for the ones completed correct?
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  2. #2
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    Quote Originally Posted by jimgreenmore View Post
    Differentiate the following
    1. (i) \frac{3}{x^2}
    I struggled with these because I'm used to differentiating square roots and fractions.
    \frac3{x^2} = 3 x^{-2}

    Now use the chain-rule.
    (ii) y = x+\sqrt{x}

    2. Find the gradient of the curve y = x^2 + \frac{1}{x} at the point (1,2)
    Answer I got = 3
    How did you get this result. Btw I've got 1.
    3. Differentiate
    (i) y=(x+2)(x^3+5x)
    Answer I got = 5x^3 + 10x - 6x^2 +10
    If I were you I would have expanded the brackets before differentiation (otherwise you have to use the product rule and the chain-rule)

    y=(x+2)(x^3+5x) = x^4+2x^3+5x^2+5x
    (ii) y=\frac{3x+2}{2x+1}
    Answer I got = \frac{7}{(2x+1)^2}
    You have to use the quotient rule (what you obviously attempted to do), but you must have made a sign mistake:

    y'=- \frac1{(2x+1)^2}

    Are my answers for the ones completed correct?
    No. See my remarks.
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  3. #3
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    Here is how I did them originally:

    Differentiate the following

    2. Find the gradient of the curve y = x^2 + \frac{1}{x} at the point (1,2)

    \frac{dy}{dx} = 2x+x^{-1}
    =2(1) + (1)^{-1}
    =2+1
    = 3

    3. Differentiate
    (i) y=(x+2)(x^3+5x)
    =(x+2)(3x^2+5)+(x^3+5x)(1)
    =4x^3+5x-6x^2+10+x^3+5x
    =5x^3 + 10x - 6x^2 +10

    (ii) y=\frac{3x+2}{2x+1}
    =\frac{dy}{dx} = 3x+2
    =\frac{dy}{dx} = 3

    =\frac{dy}{dx} = 2x+1
    =\frac{dy}{dx} = 2

    =\frac{(2x+1)3 - (3x+2)2}{(2x+1)^2}
    =\frac{(6x+3-6x+4)}{(2x+1)^2}

    =\frac{7}{(2x+1)^2}




    I did this after your help, \frac{3}{x^2}

    I got:

    = 3x^{-2}
    = -2(-3x)^{-3}(-3)
    = 6(-3x)^{-3}

    Thanks for the help so far.
    Last edited by jimgreenmore; January 28th 2010 at 01:35 PM.
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  4. #4
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    Quote Originally Posted by jimgreenmore View Post
    Here is how I did them originally:

    Differentiate the following

    2. Find the gradient of the curve y = x^2 + \frac{1}{x} at the point (1,2)

    \frac{dy}{dx} = 2x+\bold{\color{red}(-1)}x^{\bold{\color{red}-2}}
    =2(1) + \bold{\color{red}(-1)}(1)^{\bold{\color{red}-2}}
    =2\bold{\color{red}-1}
    = \bold{\color{red}1}

    3. Differentiate
    (i) y=(x+2)(x^3+5x)
    =(\bold{\color{red}x}+2)(\bold{\color{red}3x^2}+5)  +(x^3+5x)(1) <<<<<< here you calculate 1 \cdot 3 = 4. Quite unusual
    =\bold{\color{red}4x^3}+5x-6x^2+10+x^3+5x
    =5x^3 + 10x - 6x^2 +10

    (ii) y=\frac{3x+2}{2x+1}
    =\frac{dy}{dx} = 3x+2
    =\frac{dy}{dx} = 3

    =\frac{dy}{dx} = 2x+1
    =\frac{dy}{dx} = 2

    =\frac{(2x+1)3 - (3x+2)2}{(2x+1)^2}
    =\frac{(6x+3-6x\bold{\color{red}-}4)}{(2x+1)^2}

    =\frac{\bold{\color{red}-1}}{(2x+1)^2}




    I did this after your help, \frac{3}{x^2}

    I got:

    = 3 \cdot x^{-2} <<<<<<<< The number 3 is a constant factor.
    = \bold{\color{red}3 \cdot (-2)}(x)^{-3}
    ...

    Thanks for the help so far.
    ...
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  5. #5
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    I'm still lost here.
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  6. #6
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    Quote Originally Posted by jimgreenmore View Post
    (ii) y = x+\sqrt{x} at the point (1,2)
    As you know the power rule for differentiation is \frac{d}{dx}x^n = nx^{n-1}. From the laws of exponents \sqrt{x} = x^{1/2}<br />

    Therefore you get f'(x) = 1 + \frac{1}{2}x^{(1/2) - 1}

    f'(1) = 1+\frac{1}{2\sqrt1} = \frac{3}{2}



    3. Differentiate
    (ii) y=\frac{3x+2}{2x+1}
    Answer I got = \frac{7}{(2x+1)^2}

    Are my answers for the ones completed correct?
    \frac{3(2x+1)-2(3x+2)}{(2x+1)^2}

    When expanding the numerator remember that the -2 must be distributed to all terms inside the brackets:

    3(2x+1) =  6x+3
    -2(3x+2) = -6x-4

    \frac{6x+3-6x-4}{(2x+1)^2} = -\frac{1}{(2x+1)^2}
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  7. #7
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    Quote Originally Posted by e^(i*pi) View Post
    As you know the power rule for differentiation is \frac{d}{dx}x^n = nx^{n-1}. From the laws of exponents \sqrt{x} = x^{1/2}<br />

    Therefore you get f'(x) = 1 + \frac{1}{2}x^{(1/2) - 1}

    f'(1) = 1+\frac{1}{2\sqrt1} = \frac{3}{2}




    \frac{3(2x+1)-2(3x+2)}{(2x+1)^2}

    When expanding the numerator remember that the -2 must be distributed to all terms inside the brackets:

    3(2x+1) = 6x+3
    -2(3x+2) = -6x-4

    \frac{6x+3-6x-4}{(2x+1)^2} = -\frac{1}{(2x+1)^2}
    At the point (1,2) wasn't part of that first question.

    Did you mean \frac{-1}{(2x+1)^2} for the second or is -\frac{1}{(2x+1)^2} correct?
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  8. #8
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    Quote Originally Posted by jimgreenmore View Post
    At the point (1,2) wasn't part of that first question.

    Did you mean \frac{-1}{(2x+1)^2} for the second or is -\frac{1}{(2x+1)^2} correct?
    Was probably me misreading the question, ignore the f'(1) bit but the differentiation is still good.


    For the second it doesn't matter - those expressions are equal
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  9. #9
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    Oh, okay. Thanks. What is \frac{3}{x^2} differentiated?
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  10. #10
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    Quote Originally Posted by jimgreenmore View Post
    Oh, okay. Thanks. What is \frac{3}{x^2} differentiated?
    f(x) = \frac{3}{x^2}=  3x^{-2}

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