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differentiation confused?

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 dydx \displaystyle \frac{dy}{dx} use the general rule;

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

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

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

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

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

Edit: damn i am too slow with latex :angry:


I was actually thinking how much nicer it would look with latex


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Reply 4
Avatar for B_9710
B_9710
18
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).
Original post by drandy76
I was actually thinking how much nicer it would look with latex


Posted from TSR Mobile


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

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


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Reply 7
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TLDM
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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.
Reply 8
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Zacken
22
Original post by drandy76
Think the best method is a quick post before editing to put it in latex, best of both worlds


Posted from TSR Mobile


Nah, just keep using LaTeX, gets faster than not using LaTeX after a while and and some practice.
Reply 9
Avatar for kiiten
kiiten
OP
18
Original post by DylanJ42
To find dydx \displaystyle \frac{dy}{dx} use the general rule;

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

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

Edit: damn i am too slow with latex :angry:


So for x^4/4 using the indice rule do you get 4x^3 which is differentiated to get 12x^2 ?
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 f(x)=x44 \displaystyle f(x) = \frac{x^4}{4} which can be rewritten as f(x)=14⋅x4 \displaystyle f(x) = \frac{1}{4} \cdot x^4

ignore the 1/4 for now. can you differentiate x^4 for me? :biggrin:
Reply 11
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Zacken
22
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:

Unparseable latex formula:

\displaystyle [br]\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.
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.
Reply 13
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Zacken
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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\LaTeX masterrace
Original post by Zacken
LaTeX\LaTeX masterrace


I'd rep but according to TSR, you've been stealing quite a lot of them recently :colonhash:
Reply 15
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Zacken
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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 :colonhash:


Have a rep from me. :yep:
Reply 16
Avatar for kiiten
kiiten
OP
18
Original post by DylanJ42
you have f(x)=x44 \displaystyle f(x) = \frac{x^4}{4} which can be rewritten as f(x)=14⋅x4 \displaystyle f(x) = \frac{1}{4} \cdot x^4

ignore the 1/4 for now. can you differentiate x^4 for me? :biggrin:


Ahh i see - i remember this method now. Do you get x^3 ?
Original post by kiiten
Ahh i see - i remember this method now. Do you get x^3 ?


y=xny = x^{n}

dydx=nxn−1\dfrac{dy}{dx} = nx^{n-1}

in this case n=4n=4

So you are right, it would be x3x^{3} but you're missing something else!
Reply 18
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kiiten
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Original post by edothero
y=xny = x^{n}

dydx=nxn−1\dfrac{dy}{dx} = nx^{n-1}

in this case n=4n=4

So you are right, it would be x3x^{3} but you're missing something else!


Missing something? I only differentiated the first term (x^4/4) to get x^3
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 x3x^{3} when you differentiated x4x^{4}

My bad!

ddx(14x4)=x3\dfrac{d}{dx}(\dfrac{1}{4}x^{4}) = x^{3}

:biggrin::yy:
(edited 8 years ago)

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