1. ## Question about Derivative Rules

Hi, I just need clarification on the constant rule of derivatives.
The derivative of a constant = 0. So if after applying the quotient or product rule I end up with,

(x-4)d(2) + (2)d(x-4)
=(x - 4)(0) + (2)(1)
= 0 + 2

Similarly,

(x^2 + 2x + 1)d(2) + (2)d(x^2 + 2x + 1)
= (x^2 + 2x + 1)(0) + (2)(2x + 2)
= 0 + (4x + 4)

(d = derivative, i don't know how else to enter it in here.)

What I need clarification with is if we have a derivative of a constant multiplied by another function, does it always equal 0 because we multiply by 0? Or do we just remove the constant derivative?

So if we just remove the constant derivative the first equation would end up like

(x - 4) + (2x - 1) instead of 0 + (2x - 1).

2. Originally Posted by abstraktz
(x-4)d(2) + (2)d(x-4)
=(x - 4)(0) + (2)(1)
= 0 + 2
This is correct.

3. I think you could save yourself some time if you realize simply

d(cf)/dx = cdf/dx

or for someting like 2/(x+1) d (2/(x+1))/dx = 2 d(1/(x+1))/dx

4. Originally Posted by abstraktz
Hi, I just need clarification on the constant rule of derivatives.
The derivative of a constant = 0. So if after applying the quotient or product rule I end up with,

(x-4)d(2) + (2)d(x-4)
=(x - 4)(0) + (2)(1)
= 0 + 2

Similarly,

(x^2 + 2x + 1)d(2) + (2)d(x^2 + 2x + 1)
= (x^2 + 2x + 1)(0) + (2)(2x + 2)
= 0 + (4x + 4)

(d = derivative, i don't know how else to enter it in here.)

What I need clarification with is if we have a derivative of a constant multiplied by another function, does it always equal 0 because we multiply by 0? Or do we just remove the constant derivative?

So if we just remove the constant derivative the first equation would end up like

(x - 4) + (2x - 1) instead of 0 + (2x - 1).

If $y=k \: \, \: k \neq f(x)$ then $\frac{dy}{dx}(k) = 0$.

However, your question seems strange, what was the original question? You can represent derivatives with respect to x with $$\frac{d}{dx} f(x)$$. In your case replace f(x) with the function to differentiate

5. Originally Posted by abstraktz
Hi, I just need clarification on the constant rule of derivatives.
The derivative of a constant = 0. So if after applying the quotient or product rule I end up with,

(x-4)d(2) + (2)d(x-4)
=(x - 4)(0) + (2)(1)
= 0 + 2

Similarly,

(x^2 + 2x + 1)d(2) + (2)d(x^2 + 2x + 1)
= (x^2 + 2x + 1)(0) + (2)(2x + 2)
= 0 + (4x + 4)

(d = derivative, i don't know how else to enter it in here.)

What I need clarification with is if we have a derivative of a constant multiplied by another function, does it always equal 0 because we multiply by 0? Or do we just remove the constant derivative?

So if we just remove the constant derivative the first equation would end up like

(x - 4) + (2x - 1) instead of 0 + (2x - 1).

Note that $\frac{d}{dx}[kf(x)]=k\frac{d}{dx}[f(x)]$ where k is a constant. What this means is that we can pull constants out of the differentiation process.

$\frac{d}{dx}[2(x-4)]=2\frac{d}{dx}(x-4)=2(1)=2$

6. Thank you people!
Um I made that question up on the spot earlier as I didn't have it with me. But the original question was

$\frac{d}{dx} 1/x-1$

7. Originally Posted by abstraktz
Thank you people!
Um I made that question up on the spot earlier as I didn't have it with me. But the original question was

$\frac{d}{dx} 1/x-1$
$\frac{1}{x-1}$ or $\frac{1}{x}-1$?

8. $
\frac{1}{x-1}
$

which equals

$
\frac{-1}{(x-1)^2}
$

9. Originally Posted by abstraktz
$
\frac{1}{x-1}
$

which equals

$
\frac{-1}{(x-1)^2}
$
Yes, $\frac{d}{dx}\left[\frac{1}{x-1}\right]=\frac{-1}{(x-1)^2}$