# Derivative Calculation

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

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"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

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

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

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

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 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!

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 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.

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 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.

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 Humm... thanks Patrick... – Kerim Atasoy Mar 6 '12 at 21:40

Type your equation into wolfram alpha, put d/dx in front of it, click on the '=' on the right of the text entry field, wait for it to finish and finally click on 'show steps' in the top right corner of the results-box.

Alternatively click on this link, and click on the 'show steps' in the top right corner of the results-box.

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 Thanks Peter, I also would like to try "Wolfram" sites and applications, they might work to understand the subjects better... – Kerim Atasoy Mar 6 '12 at 21:44