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- #1

#### coolbeans33

##### New member

- Sep 19, 2013

- 23

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

^{3}?

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

- Thread starter coolbeans33
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- Thread starter
- #1

- Sep 19, 2013

- 23

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

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

- Admin
- #2

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?

\(\displaystyle f(x) = 2\) is

what if I want to find f'(x)=2x, or f'(x)=2x^{3}?

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

\(\displaystyle f(x) = 4 --> f'(x) = 0\) Derivative of any

If you want to derive \(\displaystyle 2x\), you now have a variable with 2x. Variables derive to

Suppose you have \(\displaystyle f(x) = 2x^3\). Multiply the

\)

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:

- Feb 21, 2013

- 739

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,

\(\displaystyle |\pi\rangle\)

- Feb 21, 2013

- 739

Just to make sure that he don't missunderstand, he forgot in the derivate \(\displaystyle x^2\) so \(\displaystyle f(x)=2x^3\) Then \(\displaystyle f'(x)=6x^2\)\(\displaystyle f(x) = 2x^3 \) --> \(\displaystyle f'(x) = 6x\)

[/MATH]

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,

\(\displaystyle |\pi\rangle\)

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

- Feb 15, 2012

- 1,967

let's find (somewhat) formally, the derivative of $f(x) = 2x^n$, and then I'll answer your first question informally.

what if I want to find f'(x)=2x, or f'(x)=2x^{3}?

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

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 $a$.