/sci/ - Science & Math

[Anonymous No.4154730]

I was just wondering, how would one differentiate <div class="math">y = |x|</div>?

[Anonymous No.4154744]

Differentiate root(x^2)

[Anonymous No.4154748]

<span class="math">y = |x| = (x^2)^{1/2}[/spoiler]
<span class="math">\displaystyle\frac{dy}{dx} = (1/2)(x^2)^{-1/2} * 2x = \frac{x}{(x^2)^{1/2}}[/spoiler]
<span class="math">\displaystyle\frac{dy}{dx} = \frac{x}{|x|}[/spoiler]

[Anonymous No.4154750]

x -> 1 if x>0, -1 if x<0
Not defined if x = 0

[Anonymous No.4154759]

>>4154748
clever. never seen someone do it like that.

[Anonymous No.4154760]

You know what's interesting? Compare x^3 with x|x|. Draw their graphs (wolfram if you like). Now look at their first and second derivatives.

[Anonymous No.4154763]

>>4154730

<span class="math">y'(x) = \frac {sqrt(x^{x})} {x}[/spoiler]

[Anonymous No.4154776]

<div class="math">\frac{dy}{dx}=H(x)-H(-x)</div>
<div class="math">\frac{d^2y}{dx^2}=2 \delta(x)</div>

You mad mathematicians?

[Anonymous No.4154796]

you don't, it isn't continuous

community college education up in this bitch

[Anonymous No.4154798]

>>4154776
At what?

[Anonymous No.4154807]

>>4154796
>discontinuous at one point
>that isn't even true anyway
>hence not differentiable anywhere

Seems legit

[Anonymous No.4154806]

>>4154796
>it isn't continuous
>community college education
Yes, this sounds like community college education.

[Anonymous No.4154808]

>>4154796
The values excluded for which f(x) is not continuous are excluded from f'(x).

[Anonymous No.4154811]

>>4154798
He is using a Dirac delta outside of an integral. Mathematicians loathe it.

[Anonymous No.4154822]

y =signx, x=/=0

[Anonymous No.4154834]

y(x) = x U(x) -x U(-x);

[Anonymous No.4154844]

<span class="math"> \displaystyle{y^2 = x} [/spoiler]

<span class="math"> \displaystyle{\frac{d}{dx}y^2 = \frac{d}{dx}x} [/spoiler]

<span class="math"> \displaystyle{2y \frac{dy}{dx} = 1} [/spoiler]

<span class="math"> \displaystyle{\frac{dy}{dx} = \frac{1}{2y}} [/spoiler]

but

<span class="math"> \displaystyle{y = \sqrt{x}} [/spoiler]

thus

<span class="math"> \displaystyle{\frac{dy}{dx} = \frac{1}{2\sqrt{x}}} [/spoiler]

[Anonymous No.4154859]

>mfw i don't understand this shit

[Anonymous No.4154910]

>>4154760
Very insteresting indeed

[Anonymous No.4154924]

>>4154811
They don't loathe it at all. To me it seems more that you have never dealt properly with distributions.

>>4154910
It's a nice example for showing student that functions whose graphs look similar don't need to have resembling derivatives in any way, in particular one can be <span class="math">C^{\infty}[/spoiler] and the other just <span class="math">C^1[/spoiler].

[Anonymous No.4154929]

>>4154859

>mfw this isn't even calc I shit