differentiating $f(x)=\sqrt{x+1}/x$ So I am not sure what I need to do to differentiate this problem. Do I use a combination of chain rule and product rule, and if so what would it look like? Thanks for the help
 A: You can use the quotient rule, or you can recognize that
$$f(x) = \frac{\sqrt{x+1}}{x} = x^{-1}\sqrt{x+1}$$
and use the product rule $(g \cdot h)' = g' \cdot h + g \cdot h'$. With $g = x^{-1}$ and $h = \sqrt{x+1}$, using the product rule you get
$$f'(x) = \frac{x^{-1}}{2\sqrt{x+1}}-x^{-2}\sqrt{x+1}$$
A: So you need to use the quotient rule http://en.wikipedia.org/wiki/Quotient_rule and also technically the chain rule but since the derivative of $x+1$ is $1$ it doesn't really matter in this case.
A: Yes, you will use the quotient rule and the chain rule. The quotient rule is:
For $\frac{f(x)}{g(x)}$, $(\frac{f(x)}{g(x)})'$ = $\frac{g(x)(f(x))'-f(x)(g(x))'}{(g(x))^{2}}$. 
A: Quotient Rule derives the fraction. Chain rule derives the $\sqrt{x+1}$
The quotient rule (if you dont know it) is : $$\frac{d}{{dx}}\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \frac{{\frac{d}{{dx}}f\left( x \right)g\left( x \right) - f\left( x \right)\frac{d}{{dx}}g\left( x \right)}}{{g^2 \left( x \right)}}$$
