# How do you find the derivative of (sqrt(x^2-x+1)-1)/(1+ sqrtx)?

May 21, 2015

You find it very carefully...lol

Use the Quotient Rule, Power Rule, Linearity, and the Chain Rule:

$\frac{d}{\mathrm{dx}} \left(\setminus \frac{\setminus \sqrt{{x}^{2} - x + 1} - 1}{1 + \setminus \sqrt{x}}\right)$

$= \setminus \frac{\left(1 + \setminus \sqrt{x}\right) \setminus \cdot \setminus \frac{1}{2} {\left({x}^{2} - x + 1\right)}^{- \frac{1}{2}} \setminus \cdot \left(2 x - 1\right) - \left(\setminus \sqrt{{x}^{2} - x + 1} - 1\right) \setminus \cdot \setminus \frac{1}{2} {x}^{- \frac{1}{2}}}{{\left(1 + \setminus \sqrt{x}\right)}^{2}}$

$= \setminus \frac{\setminus \frac{2 x - 1 + 2 {x}^{\frac{3}{2}} - {x}^{\frac{1}{2}}}{2 \sqrt{{x}^{2} - x + 1}} - \setminus \frac{\sqrt{{x}^{2} - x + 1} - 1}{2 \sqrt{x}}}{{\left(1 + \setminus \sqrt{x}\right)}^{2}}$

$= \setminus \frac{2 {x}^{\frac{3}{2}} - {x}^{\frac{1}{2}} + 2 {x}^{2} - x - \left({x}^{2} - x + 1 - \sqrt{{x}^{2} - x + 1}\right)}{2 \sqrt{x} \sqrt{{x}^{2} - x + 1} {\left(1 + \sqrt{x}\right)}^{2}}$

$= \setminus \frac{{x}^{2} + 2 x \sqrt{x} - \sqrt{x} - 1 + \sqrt{{x}^{2} - x + 1}}{2 \sqrt{{x}^{3} - {x}^{2} + x} {\left(1 + \sqrt{x}\right)}^{2}}$