Answer to Question #257106 in Calculus for Sid

Question

use chain rule to find dy/dx for given value of x

y=(u-1/u+1)^1/2, u=√x-1, for x=34/9

Answer 1

$$y=(\dfrac{u-1}{u+1})^{1/2}, u=\sqrt{x-1}$$

$$y'_x=\dfrac{1}{2}(\dfrac{u-1}{u+1})^{-1/2}(\dfrac{u+1-(u-1)}{(u+1)^2})u'_x$$

$$=(\dfrac{u+1}{u-1})^{1/2}(\dfrac{1}{(u+1)^2})(\dfrac{1}{2\sqrt{x-1}})$$

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

$$x=\dfrac{34}{9}:u=\sqrt{\dfrac{34}{9}-1}=\dfrac{5}{3}$$

$$y'(\dfrac{34}{9})=\bigg(\dfrac{5/3+1}{5/3-1}\bigg)^{1/2}\big(\dfrac{1}{(5/3+1)^2}\big)(\dfrac{1}{2(5/3})$$

$$=\dfrac{27}{320}$$