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    If you do dy/dx of x^4/4 + 32/x^2 does it become dy/dx of 4x^4-1 + 32x-2 ?
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    X^3 -64x^-3 is what you should get, remember the power of X is multiplied by the coefficient and and the new power is n-1 such that: if y = x^n
    Dy/dx = nx^n-1


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    (Original post by kiiten)
    If you do dy/dx of x^4/4 + 32/x^2 does it become dy/dx of 4x^4-1 + 32x-2 ?
    To find  \displaystyle \frac{dy}{dx} use the general rule;

    if  \displaystyle y = x^n then  \displaystyle \frac{dy}{dx} = n \cdot x^{n-1}

    For the second term dont forget to use the indice rule  \displaystyle \frac{1}{a^n} = a^{-n}

    Edit: damn i am too slow with latex
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    (Original post by DylanJ42)
    To find  \displaystyle \frac{dy}{dx} use the general rule;

    if  \displaystyle y = x^n then  \displaystyle \frac{dy}{dx} = n \cdot x^{n-1}

    For the second term dont forget to use the indice rule  \displaystyle \frac{1}{a^n} = a^{-n}

    Edit: damn i am too slow with latex
    I was actually thinking how much nicer it would look with latex


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    (Original post by kiiten)
    If you do dy/dx of x^4/4 + 32/x^2 does it become dy/dx of 4x^4-1 + 32x-2 ?
    You can't do dy/dx of some function or expression. You do d/dx of something. If you have y=f(x) and differentiate both sides wrt x, you get dy/dx=f'(x).
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    (Original post by drandy76)
    I was actually thinking how much nicer it would look with latex


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    and im thinking of how much quicker the reply would be without latex :laugh:

    we can never win
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    Think the best method is a quick post before editing to put it in latex, best of both worlds


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    I try to do things in LaTeX first. I find it hard to understand things not formatted nicely. But yeah, it is slow. I need to practice more so I don't have to keep looking at the wiki page.
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    (Original post by drandy76)
    Think the best method is a quick post before editing to put it in latex, best of both worlds


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    Nah, just keep using LaTeX, gets faster than not using LaTeX after a while and and some practice.
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    (Original post by DylanJ42)
    To find  \displaystyle \frac{dy}{dx} use the general rule;

    if  \displaystyle y = x^n then  \displaystyle \frac{dy}{dx} = n \cdot x^{n-1}

    For the second term dont forget to use the indice rule  \displaystyle \frac{1}{a^n} = a^{-n}

    Edit: damn i am too slow with latex
    So for x^4/4 using the indice rule do you get 4x^3 which is differentiated to get 12x^2 ?
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    (Original post by kiiten)
    So for x^4/4 using the indice rule do you get 4x^3 which is differentiated to get 12x^2 ?
    you have  \displaystyle f(x) = \frac{x^4}{4} which can be rewritten as  \displaystyle  f(x) = \frac{1}{4} \cdot x^4

    ignore the 1/4 for now. can you differentiate x^4 for me?
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    (Original post by kiiten)
    So for x^4/4 using the indice rule do you get 4x^3 which is differentiated to get 12x^2 ?
    Differentiating is a liner operation. i.e:

    \displaystyle 

\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}x} \left(af(x)\right) = a\frac{\mathrm{d}}{\mathrm{d}x} \left(f(x)\right)\end{equation*}

    Use this in conjunction with DylanJ's post above.
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    (Original post by Zacken)
    Nah, just keep using LaTeX, gets faster than not using LaTeX after a while and and some practice.
    This, I always use LaTeX and I've become quite good at it now. I can type up things quite fast.
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    (Original post by edothero)
    This, I always use LaTeX and I've become quite good at it now. I can type up things quite fast.
    \LaTeX masterrace
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    (Original post by Zacken)
    \LaTeX masterrace
    I'd rep but according to TSR, you've been stealing quite a lot of them recently
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    (Original post by edothero)
    I'd rep but according to TSR, you've been stealing quite a lot of them recently
    Have a rep from me. :yep:
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    (Original post by DylanJ42)
    you have  \displaystyle f(x) = \frac{x^4}{4} which can be rewritten as  \displaystyle  f(x) = \frac{1}{4} \cdot x^4

    ignore the 1/4 for now. can you differentiate x^4 for me?
    Ahh i see - i remember this method now. Do you get x^3 ?
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    (Original post by kiiten)
    Ahh i see - i remember this method now. Do you get x^3 ?
    y = x^{n}

    \dfrac{dy}{dx} = nx^{n-1}

    in this case n=4

    So you are right, it would be x^{3} but you're missing something else!
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    (Original post by edothero)
    y = x^{n}

    \dfrac{dy}{dx} = nx^{n-1}

    in this case n=4

    So you are right, it would be x^{3} but you're missing something else!
    Missing something? I only differentiated the first term (x^4/4) to get x^3
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    (Original post by kiiten)
    Missing something? I only differentiated the first term (x^4/4) to get x^3
    Oh I see, no you're correct! I thought you meant you got x^{3} when you differentiated x^{4}

    My bad!

    \dfrac{d}{dx}(\dfrac{1}{4}x^{4}) = x^{3}

    :yy:
 
 
 
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