Question #257106

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


1
Expert's answer
2021-10-27T12:21:23-0400
y=(u1u+1)1/2,u=x1y=(\dfrac{u-1}{u+1})^{1/2}, u=\sqrt{x-1}

yx=12(u1u+1)1/2(u+1(u1)(u+1)2)uxy'_x=\dfrac{1}{2}(\dfrac{u-1}{u+1})^{-1/2}(\dfrac{u+1-(u-1)}{(u+1)^2})u'_x

=(u+1u1)1/2(1(u+1)2)(12x1)=(\dfrac{u+1}{u-1})^{1/2}(\dfrac{1}{(u+1)^2})(\dfrac{1}{2\sqrt{x-1}})

=(x1+1x11)1/2(1(x1+1)2)(12x1)=(\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=349:u=3491=53x=\dfrac{34}{9}:u=\sqrt{\dfrac{34}{9}-1}=\dfrac{5}{3}

y(349)=(5/3+15/31)1/2(1(5/3+1)2)(12(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})

=27320=\dfrac{27}{320}



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