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    How do I differentiate y=[(x+3)^4]-4x
    I know how to use the chain rule, but what do I do with the -4x bit? If t=x+3, should i differentiate y=(t^3)-4x?
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    \displaystyle \frac{\mathrm{d}}{\mathrm{d}x}y = \frac{\mathrm{d}}{\mathrm{d}x} \bigg( (x+3)^4 \bigg) - \frac{\mathrm{d}}{\mathrm{d}x}(4  x).

    The differential operator is linear, the derivative of a sum of function is the sum of the derivatives of each individual function.
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    (Original post by yoshimatey)
    How do I differentiate y=[(x+3)^4]-4x
    I know how to use the chain rule, but what do I do with the -4x bit? If t=x+3, should i differentiate y=(t^3)-4x?
    This video should explain everything:
    http://www.examsolutions.net/maths-r...tutorial-1.php
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    Welcome Squad
    (Original post by yoshimatey)
    How do I differentiate y=[(x+3)^4]-4x
    I know how to use the chain rule, but what do I do with the -4x bit? If t=x+3, should i differentiate y=(t^3)-4x?
    the -4x is a separate term
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    (Original post by yoshimatey)
    How do I differentiate y=[(x+3)^4]-4x
    I know how to use the chain rule, but what do I do with the -4x bit? If t=x+3, should i differentiate y=(t^3)-4x?
    The chain rule, (d/Dx)((x+3)^4)=(du^4)/du) (du/dx) where u=x+3 and (d/du) u^4= 4u^3


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    just differentiate each bit separately ?
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    (Original post by Zacken)
    \displaystyle \frac{\mathrm{d}}{\mathrm{d}x}y = \frac{\mathrm{d}}{\mathrm{d}x} \bigg( (x+3)^4 \bigg) - \frac{\mathrm{d}}{\mathrm{d}x}(4  x).

    The differential operator is linear, the derivative of a sum of function is the sum of the derivatives of each individual function.
    THANKS A BILLION! You have really helped me out!
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    (Original post by yoshimatey)
    THANKS A BILLION! You have really helped me out!
    You're very welcome!

    You can even check this work for certain functions y = 3x = x + x + x, so \frac{dy}{dx} = \frac{\mathrm{d}}{\mathrm{d}x}x + \frac{\mathrm{d}}{\mathrm{d}x}x + \frac{\mathrm{d}}{\mathrm{d}x}x = 1 + 1 + 1 = 3

    But also, \frac{\mathrm{d}}{\mathrm{d}x}y = \frac{\mathrm{d}}{\mathrm{d}x}(3  x) = 3
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    (Original post by Zacken)
    You're very welcome!

    You can even check this work for certain functions y = 3x = x + x + x, so \frac{dy}{dx} = \frac{\mathrm{d}}{\mathrm{d}x}x + \frac{\mathrm{d}}{\mathrm{d}x}x + \frac{\mathrm{d}}{\mathrm{d}x}x = 1 + 1 + 1 = 3

    But also, \frac{\mathrm{d}}{\mathrm{d}x}y = \frac{\mathrm{d}}{\mathrm{d}x}(3  x) = 3
    Or prove it.
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    (Original post by morgan8002)
    Or prove it.
    Indeed, although it might be a little soon to ask the person to prove it given their unfamiliarity with limits, but for anybody interested:

    Let h(x) = f(x) + g(x), then

    \displaystyle h'(x) = \lim_{a \to 0} \frac{h(x+a) - h(x)}{a}

    \displaystyle = \lim_{a \to 0}\frac{f(x+a) + g(x+a) - f(x) - g(x)}{a}

    and a bit of shuffling and the sum rule for limits yields


    \displaystyle h'(x) = \lim_{a \to 0} \frac{f(x+a) - f(x)}{a} + \lim_{a \to 0}\frac{g(x+a) - g(x)}{a} = f'(x) + g'(x)
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    (Original post by Zacken)
    Indeed, although it might be a little soon to ask the person to prove it given their unfamiliarity with limits, but for anybody interested:

    Let h(x) = f(x) + g(x), then

    \displaystyle h'(x) = \lim_{a \to 0} \frac{h(x+a) - h(x)}{a}

    \displaystyle = \lim_{a \to 0}\frac{f(x+a) + g(x+a) - f(x) - g(x)}{a}

    and a bit of shuffling and the sum rule for limits yields


    \displaystyle h'(x) = \lim_{a \to 0} \frac{f(x+a) - f(x)}{a} + \lim_{a \to 0}\frac{g(x+a) - g(x)}{a} = f'(x) + g'(x)
    'overkill mate' - Ronald Weasley
 
 
 
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