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#### coolbeans33

##### New member
I don't understand this concept very well. So suppose I want to find the derivative of the function f(x)=2. is the answer just zero?

what if I want to find f'(x)=2x, or f'(x)=2x3?

please explain how to get the answer step by step, is there some equation I use or something? I just don't get it!

#### MarkFL

Staff member
Yes, the derivative of a constant is zero...because a constant does not change.

Let me ask you, are you familiar with the definition:

$$\displaystyle f'(x)\equiv\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$ ?

It is from this definition that the rules of differentiation are derived. Have you used this "first principle" to differentiate simple functions?

#### shamieh

##### Active member
I don't understand this concept very well. So suppose I want to find the derivative of the function f(x)=2. is the answer just zero?

what if I want to find f'(x)=2x, or f'(x)=2x3?

please explain how to get the answer step by step, is there some equation I use or something? I just don't get it!
$$\displaystyle f(x) = 2$$ is always $$\displaystyle f'(x) = 0$$.. -->$$\displaystyle f(x) = 3$$ --> $$\displaystyle f'(x) = 0$$

$$\displaystyle f(x) = 4 --> f'(x) = 0$$ Derivative of any number goes to zero. (Read the chapter in your book over derivatives.)

If you want to derive $$\displaystyle 2x$$, you now have a variable with 2x. Variables derive to 1. so if you had $$\displaystyle f(x) = x$$, then $$\displaystyle f'(x) = 1$$. So, if you have $$\displaystyle f(x) = 2x$$, then $$\displaystyle f'(x) = 2$$. because $$\displaystyle 2$$(derivative of x, which is 1) = $$\displaystyle 2$$. (Because $$\displaystyle (2)(1)=2$$.)

Suppose you have $$\displaystyle f(x) = 2x^3$$. Multiply the exponential by the base, then take the exponential and subtract 1. $$\displaystyle f(x) = 2x^3$$ --> $$\displaystyle f'(x) = 6x^2$$ Here is another example $$\displaystyle f(x) = 4x^3--> f'(x) = 12x^2$$

One More.
$$\displaystyle f(x) = 100x^{1000}$$
$$\displaystyle f'(x) = 100000x^{999}$$

Of course I have explained it in the simplest of terms. MarkFL and the rest will be able to expand on derivatives in a more technical and detailed way. Also remember

$$\displaystyle f(x) = \pi^2$$
$$\displaystyle f'(x) = 0.$$

Last edited:

#### Petrus

##### Well-known member
Hello,
Or you could rewrite $$\displaystyle f(x)=2$$ as $$\displaystyle f(x)=2x^0$$ right? Cause $$\displaystyle x^0=1$$. and if we derivate we get $$\displaystyle f'(x)=0•2x^{-1} <=> f'(x)=0$$. Depends if you read in university or high school.. If you read in university you should check look at MarkFL comment and we can help you if you need to understand!

Regards,
$$\displaystyle |\pi\rangle$$

#### Petrus

##### Well-known member
$$\displaystyle f(x) = 2x^3$$ --> $$\displaystyle f'(x) = 6x$$
[/MATH]
Just to make sure that he don't missunderstand, he forgot in the derivate $$\displaystyle x^2$$ so $$\displaystyle f(x)=2x^3$$ Then $$\displaystyle f'(x)=6x^2$$
Or lets say c is a constant $$\displaystyle f(x)=c^n$$ Then $$\displaystyle f'(x)=n•c^{n-1}$$
BUT there is some rule for like derivate $$\displaystyle f(x)g( x)$$ but I asume that you have not learned that just to make sure that you think derivate works like that always Regards,
$$\displaystyle |\pi\rangle$$

#### shamieh

##### Active member
Just to make sure that he don't missunderstand, he forgot in the derivate $$\displaystyle x^2$$ so $$\displaystyle f(x)=2x^3$$ Then $$\displaystyle f'(x)=6x^2$$
Exactly, typo on my part. Lol good clarification.

#### Deveno

##### Well-known member
MHB Math Scholar
I don't understand this concept very well. So suppose I want to find the derivative of the function f(x)=2. is the answer just zero?

what if I want to find f'(x)=2x, or f'(x)=2x3?

please explain how to get the answer step by step, is there some equation I use or something? I just don't get it!
let's find (somewhat) formally, the derivative of $f(x) = 2x^n$, and then I'll answer your first question informally.

By definition:

$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim_{h \to 0} \frac{2(x+h)^n -2 x^n}{h}$

$\displaystyle = \lim_{h \to 0}\frac{2x^n + 2nx^{n-1}h + \cdots + 2nxh^{n-1} + 2h^n - 2x^n}{h}$

$\displaystyle = \lim_{h \to 0} \frac{2h(nx^{n-1} + \cdots + nxh^{n-2} + h^{n-1})}{h}$

(all the terms in the "..." part contain a factor of $h$)

$\displaystyle = \lim_{h \to 0} 2nx^{n-1} + h(\text{other terms})$

$= 2nx^{n-1} + 0\ast(\text{who cares?}) = 2nx^{n-1}$.

However, for $n = 0$ there is an easier way:

If $f(x) = 2$ (for ALL $x$), then:

$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

$\displaystyle = \lim_{h \to 0} \frac{2 - 2}{h} = \lim_{h \to 0} \frac{0}{h} = 0$.

*************

Informally, the derivative $f'(a)$ measures the rate of change (or SLOPE) of the function $f$ at the point $a$. If $f$ is a constant function, it never changes, it remains constant, so its rate of change is 0 (no change) at every point $a$.