Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've some equation about "Derivatives" to ask about. Please, show me how to do that step by step: $$f(x) = \frac{3x^2+1}{2}.$$ $f'(x)= ?$

share|cite|improve this question
"you people" can seem somewhat dismissive or derogatory, so I took the liberty of changing it. – Arturo Magidin Mar 6 '12 at 20:37
Do you know all the properties of the operation, "taking derivative" ? – user21436 Mar 6 '12 at 20:38
Yes sir, it's " f`(x)= ? " – Kerim Atasoy Mar 6 '12 at 20:38
What do you know about derivatives? Do you know/are you allowed to use some (which?) basic properties of derivatives? Some formulas? Do you need to compute $f'(x)$ using the limit definition? – Arturo Magidin Mar 6 '12 at 20:38
@KannappanSampath: I've been studying "Derivatives" since last morning, so, I don't know that much about it for now... Any help, please? – Kerim Atasoy Mar 6 '12 at 20:41
up vote 2 down vote accepted

$\frac{1}{2}\frac{d}{dx}(3x^2 + 1) = \frac{1}{2}(\frac{d}{dx}3x^2 + \frac{d}{dx}1) = \frac{1}{2}(3\frac{d}{dx}x^2 + \frac{d}{dx}1)$

Since 1 is a constant its derivative becomes 0 and as for $x^2$ we have a rule that states that if $f(x) = x^r$ then $f'(x) = r\cdot x^{r-1}$. With that in mind we get

$\frac{1}{2}(3\frac{d}{dx}x^2 + \frac{d}{dx}1) = \frac{1}{2}(3(2x) + 0) = \frac{1}{2}6x = 3x$

share|cite|improve this answer
I understand it now, thanks sir... :) – Kerim Atasoy Mar 6 '12 at 21:29

Each step should follow one of the derivative rules that you know about. The notation "$\frac{d}{dx}$" in what follows means "take the derivative of".

$$ \begin{align*} f'(x) &= \frac{d}{dx}\left[ \frac{3x^2 + 1}{2} \right]\\ &= \frac{d}{dx}\left[ \frac{1}{2}(3x^2 + 1) \right], \quad\textrm{(algebra)}\\ &= \frac{1}{2}\frac{d}{dx}\left[3x^2 + 1 \right], \quad\textrm{(constant multiple rule)}\\ &= \frac{1}{2}\left( \frac{d}{dx}[3x^2] + \frac{d}{dx}[1] \right), \quad \textrm{(sum/difference rule)}\\ &= \frac{1}{2}\left( 3\frac{d}{dx}[x^2] + \frac{d}{dx}[1] \right), \quad \textrm{(constant mult. rule again)}\\ &= \frac{1}{2}\left( 3(2x) + \frac{d}{dx}[1] \right), \quad \textrm{(power rule)}\\ &= \frac{1}{2}\left( 3(2x) + 0 \right), \quad \textrm{(derivative of a constant is 0 -- really just power rule)}\\ &= 3x, \quad \textrm{(algebra to simplify answer)} \end{align*} $$ Now as you do more and more of these problems, you'll find which steps you can do in your head, until you get to the point where it becomes a one-line problem!

Hope this helps!

share|cite|improve this answer
Thanks @Arturo. I just corrected it – Shaun Ault Mar 6 '12 at 20:55
An other awesome help here, thanks Shaun... :) – Kerim Atasoy Mar 6 '12 at 21:32

It's simple: Just apply the definition of the derivative ($f$ is a polynomial so is differentiable, which we can prove).

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

$\lim \limits_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim \limits_{h \to 0} \frac{(3(x+h)^2 + 1) -(3x^2 +1)}{2h} = $ . . . ?

All it takes is a little manipulation. You should find some very important things will cancel out and the limit will be easy to take.

share|cite|improve this answer
Well, I'm not skilled enough to understand your very valuable reply here for now, but thank you very much Tyler... :) This might work for others at least... :) – Kerim Atasoy Mar 6 '12 at 21:37

$f(x) = \frac32 x^2 + \frac12$. Just use the Power Rule.

share|cite|improve this answer
Humm... thanks Patrick... – Kerim Atasoy Mar 6 '12 at 21:40

Easy:$\frac{d}{dx} {\frac{3}{2}x^{2}+\frac{1}{2}} = 3x.$ Just do some algebra, and then use the Sum Rule: $\frac{d(f+g)}{dx} = \frac{df}{dx} + \frac{dg}{dx}$ followed by the Power Rule: $\frac{d}{dx} x^{n} = nx^{n-1}$. Remember that you don't have to differentiate coefficients, just leave them out of the derivative.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.