Answer to Question #313844 in Calculus for Muncher

Question

y=(x-1)√x²-2x+2

Answer 1

Solution

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

$\frac{dy}{dx}=(x-1)\frac{d}{dx}\sqrt{x^2-2x+2}+sqrt{x^2-2x+2}\frac{d}{dx}(x-1)$

$\frac{dy}{dx}=(x-1)\frac{1}{2}(x^2-2x+2)^{-\frac{1}{2}}\frac{d}{dx}(x^2-2x+2)+\sqrt{x^2-2x+2}(1-0)$

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

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

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

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

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

$\frac{dy}{dx}=\frac{(2x^2-4x+3)}{\sqrt{x^2-2x+2}}$