differentiation question!

**jvignacio** · September 6th 2008, 01:46 AM

$ f(x) = x\sqrt{x^2+1} $

what i have so far is, leaving the x on the left aside for abit.

$ f(x) = \sqrt{x^2+1} = (x^2 + 1)^\frac{1}{2} $

then

$ f(x) = \frac{1}{2}(x^2 + 1)^\frac{-1}{2} = \frac{1}{2\sqrt{x^2 + 1}} $

then

$ f(x) = \frac{1}{2\sqrt{x^2 + 1}} \cdot (2x) = \frac{2x}{2\sqrt{x^2 + 1}} $

now what do i do with the x? (thats if my calculations are correct)

**dammitpoo** · September 6th 2008, 02:01 AM

Product rule:
$ \frac{d}{dx}\ (u\cdot v) = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} $

**woody198403** · September 6th 2008, 02:14 AM

To continue from what dammitpoo said

the product rule is

$\frac{d}{dx}[f(x)g(x)] = f(x) * \frac{d}{dx}[g(x)] + g(x)* \frac{d}{dx}[f(x)]$

where $f(x) = x$ and $g(x) = \sqrt{x^2 + 1}$

So $\frac{d}{dx}(f(x)) = 1$ and $\frac{d}{dx}g(x) = \frac{x}{\sqrt{x^2 + 1}}$ ............which is what you calculated but you forgot to cancel the 2's.

You should be able to finish it from here....just plug it in to the product rule above.

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