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

1 Answer
May 21, 2015

You find it very carefully...lol

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

#d/dx(\frac{\sqrt{x^2-x+1}-1}{1+\sqrt{x}})#

#=\frac{(1+\sqrt{x})\cdot \frac{1}{2}(x^{2}-x+1)^{-1/2}\cdot (2x-1)-(\sqrt{x^{2}-x+1}-1)\cdot \frac{1}{2}x^{-1/2}}{(1+\sqrt{x})^{2}}#

#=\frac{\frac{2x-1+2x^{3/2}-x^{1/2}}{2sqrt{x^2-x+1}}-\frac{sqrt(x^2-x+1}-1}{2sqrt{x}}}{(1+\sqrt{x})^2}#

#=\frac{2x^{3/2}-x^{1/2}+2x^2-x-(x^2-x+1-sqrt{x^2-x+1})}{2sqrt{x}sqrt{x^2-x+1}(1+sqrt{x})^2}#

#=\frac{x^2+2xsqrt{x}-sqrt{x}-1+sqrt{x^2-x+1}}{2sqrt{x^3-x^2+x}(1+sqrt{x})^2}#