# differentiation question!

• Sep 6th 2008, 01:46 AM
jvignacio
differentiation question!
$\displaystyle f(x) = x\sqrt{x^2+1}$

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

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

then

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

then

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

now what do i do with the x? (thats if my calculations are correct)
• Sep 6th 2008, 02:01 AM
dammitpoo
Product rule:
$\displaystyle \frac{d}{dx}\ (u\cdot v) = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx}$
• Sep 6th 2008, 02:14 AM
woody198403
To continue from what dammitpoo said

the product rule is

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

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

So $\displaystyle \frac{d}{dx}(f(x)) = 1$ and $\displaystyle \frac{d}{dx}g(x) = \frac{x}{\sqrt{x^2 + 1}}$............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.