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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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    :) Thank you guys, now I'm having a litte affair to choose more prior, maybe a better answer from all these replies... :) Thanks...2012-03-06

5 Answers 5

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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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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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    An other awesome help here, thanks Shaun... :)2012-03-06
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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... :)2012-03-06
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$\frac{1}{2}\frac{d}{dx}(3x^2 + 1) = \frac{1}{2}(\frac{d}{dx}3x^2 + \frac{d}{dx}1) = \frac{1}{2}(3\frac{d}{dx}x^2 + \frac{d}{dx}1)$

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... :)2012-03-06
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$f(x) = \frac32 x^2 + \frac12$. Just use the Power Rule.

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    Humm... thanks Patrick...2012-03-06