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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
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5 Answers

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

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

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I understand it now, thanks sir... :) –  Kerim Atasoy Mar 6 '12 at 21:29
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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
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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
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$f(x) = \frac32 x^2 + \frac12$. Just use the Power Rule.

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

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