MHB How is the constant '4' handled in differentiation of $y=4\cos(x^3-2x)$?

[DeusAbscondus]

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]

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

 

[soroban]

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 y \:=\:c\!\cdot\!f(x), then \tfrac{dy}{dx} \:=\:c\!\cdot\!f'(x)If you wish, you can use the Product Rule on: y \:=\:5\!\cdot\!x^3

. . \frac{dy}{dx} \:=\:0\!\cdot\!x^3 + 5\!\cdot\!3x^2 \:=\:15x^2I think you'll agree that it's a waste of time.

 

[DeusAbscondus]

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]

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]

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]

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]

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]

I did not assume $y=f(x)$ to make it more explicit,

It's interchangeable in the equation.

 

[DeusAbscondus]

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