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[SOLVED] Contant included in derivative

DeusAbscondus

Active member
Jun 30, 2012
176
In the same vein as last question:

by what rule of differentiation does the constant '4' in the following expression, get included, unmodified, in the differentiation

$$y=4cos(x^3-2x) \Rightarrow y'=-4(3x^2-2).sin(x^3-2x)$$

A cross-wire in my mind wants to apply the rule:
$$y=c\Rightarrow y'=0, where, c=constant$$

Incidentally, I've searched and searched and can't find an explanation as to the logical/operational difference between the following:
$\Rightarrow$ and $\rightarrow$
Please explain,
thanks again (hellow Sudharaka)
DeusAbs
 

pickslides

Member
Feb 1, 2012
57
$$ \frac{d}{dx} (c \times f(x)) = c \frac{d}{dx}(f(x))$$
 

soroban

Well-known member
Feb 2, 2012
409
Hello, DeusAbscondus!


By what rule of differentiation does the constant '4' in the following expression,
get included, unmodified, in the differentiation

$y=4cos(x^3-2x) \Rightarrow y'=-4(3x^2-2)\sin(x^3-2x)$

One of the basic differentiation rules says:

If [tex]y \:=\:c\!\cdot\!f(x)[/tex], then [tex]\tfrac{dy}{dx} \:=\:c\!\cdot\!f'(x)[/tex]


If you wish, you can use the Product Rule on: [tex]y \:=\:5\!\cdot\!x^3[/tex]

. . [tex]\frac{dy}{dx} \:=\:0\!\cdot\!x^3 + 5\!\cdot\!3x^2 \:=\:15x^2[/tex]


I think you'll agree that it's a waste of time.
 

DeusAbscondus

Active member
Jun 30, 2012
176
Re: Contant/derivative: And what is the difference between the d/dx and dy/dx?

$$ \frac{d}{dx} (c \times f(x)) = c \frac{d}{dx}(f(x))$$
Thanks PickSlides:
just to be sure I've understood:
Is that to say $$\frac{dy}{dx}(c\times f(x))=c\frac{dy}{dx}(f(x))$$

And hello, incidentally, from Lismore, NSW.

DeusAbs
 

pickslides

Member
Feb 1, 2012
57
Yes, just ignore the constant term in the derivative and add it back in later.

I'm from Gowanbrae, VIC. Nice to meet ya!
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Deus, I'd just be more careful about the notation. Notice that pickslides wrote

$$\frac{d}{dx} (c \times f(x)) = c \frac{d}{dx}(f(x))$$

whereas you wrote

$$\frac{dy}{dx}(c\times f(x))=c\frac{dy}{dx}(f(x)).$$
There's a not so subtle difference: pickslides differentiated with respect to $x$, what you did was essentially multiply the function $c f(x)$ by a derivative $\frac{dy}{dx}$. It's not what you meant because of the context, but watch for how you write it. :)

Cheers.
 

DeusAbscondus

Active member
Jun 30, 2012
176
Deus, I'd just be more careful about the notation. Notice that pickslides wrote

$$\frac{d}{dx} (c \times f(x)) = c \frac{d}{dx}(f(x))$$

whereas you wrote

$$\frac{dy}{dx}(c\times f(x))=c\frac{dy}{dx}(f(x)).$$
There's a not so subtle difference: pickslides differentiated with respect to $x$, what you did was essentially multiply the function $c f(x)$ by a derivative $\frac{dy}{dx}$. It's not what you meant because of the context, but watch for how you write it. :)

Cheers.
Thanks Fantini. But please, when you have a moment, explain the difference between the two notations more explicitly.
Appreciated.
DeusAbs
(I never dreamed when I hit upon my dog-latin nick, Deus Abscondus: ie: MissingGod or, God is Missing, that I would end up being addressed as Deus!)
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
The first is applying the differentiation with respect to $x$ operator, whereas the second is the derivative of a function $y$ with respect to x being multiplied by $c f(x)$. Of course, I did not assume $y=f(x)$ to make it more explicit, and even if you do it doesn't change anything.

Also, I just went for something short (Thinking). I can call you Abs from now on.

Cheers.
 

pickslides

Member
Feb 1, 2012
57

DeusAbscondus

Active member
Jun 30, 2012
176
It's interchangeable in the equation.
That's a relief: any more "things I don't understand" might have proven too much for my weakened self-confidence, which has taken a beating this week, so far.

Thanks mate,
DeusAbs