If y=x^3, prove that x(dy/dx)=3y for all values of x.

any help is really appreciated, thx

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- Apr 23rd 2013, 12:19 AMJellyOnionDerivative proof
If y=x^3, prove that x(dy/dx)=3y for all values of x.

any help is really appreciated, thx - Apr 23rd 2013, 01:30 AMagentmulderRe: Derivative proof
$\displaystyle \frac{d}{dx}(y) = \frac{d}{dx}(x^3) $

$\displaystyle \frac{dy}{dx} = 3x^2 $

multiply both sides by x

$\displaystyle x \frac{dy}{dx} = 3x^3 $

but y = x^3 so substitute

$\displaystyle x \frac{dy}{dx} = 3y $

I'm not sure about the 'all values of x' part. If x = 0 then we are zeroing out both sides of the equation and that's usually a no-no.

On the other hand , the equation was true for all real values of x BEFORE i multiplied by x so even if x = 0 we get 0 = 0 which is still true.

Maybe other MHF members can provide some insightful input regarding this matter.

:) - Apr 23rd 2013, 03:07 AMJellyOnionRe: Derivative proof
Thx agentmulder

- Apr 23rd 2013, 03:25 AMzhandeleRe: Derivative proof
It works at x=0 too. Plug in x=0 y=(0)^3 it works. The continuity of x^3 guarantees it. You could make a limit argument, as x goes to 0.

- Apr 24th 2013, 11:32 PMhollywoodRe: Derivative proof
There is usually no problem with multiplying both sides of an equation by an expression that might be zero. It's when we divide by an expression that might be zero that we run into problems. For example, if x=2 and y=3,

x-1 = y-2 is true, and if we multiply both sides by (x-2), we get (x-2)(x-1) = (x-2)(y-2) which is also true. But:

(x-2)(y-2) = (x-2)(x-3) is true, and if we divide both sides by (x-2) we get y-2 = x-3, which is not true.

- Hollywood