How do you find the derivative of #F(x) = sqrt( (x-8)/(x^2-2) )#?

2 Answers
Oct 3, 2017

#F(x)=sqrt((x-8)/(x^2-2))#
#F'(x)=1/(2sqrt((x-8)/(x^2-2)))(((x^2-2)d/dx(x-8)-(x-8)d/dx(x^2-2))/(x^2-2)^2)#

#F'(x)=1/(2sqrt((x-8)/(x^2-2)))(((x^2-2)(1)-(x-8)(2x))/(x^2-2)^2)#

#F'(x)=sqrt(x^2-2)/(2sqrt((x-8)))((x^2-2-2x^2+16x)/(x^2-2)^2)#

#F'(x)=sqrt(x^2-2)/(2sqrt((x-8)))((-x^2+16x-2)/(x^2-2)^2)#

Oct 3, 2017

Answer:

# F'(x)=-((x^2-16x+2))/{2(x-8)^(1/2)(x^2-2)^(3/2)}.#

Explanation:

Let, #y=F(x)=sqrt{(x-8)/(x^2-2)}={(x-8)/(x^2-2)}^(1/2).#

#:. lny=1/2ln((x-8)/(x^2-2))=1/2{ln(x-8)-ln(x^2-2)}.#

#:. d/dx{lny}=1/2d/dx{ln(x-8)-ln(x^2-2)}.#

By the Chain Rule, then, we have,

#:. d/dy(lny)*dy/dx#

#=1/2[1/(x-8)*d/dx(x-8)-1/(x^2-2)*d/dx(x^2-2)}.#

#:. 1/y*dy/dx=1/2{1/(x-8)*1-1/(x^2-2)*2x},#

#=1/2{{(x^2-2)-2x(x-8)}/{(x-8)(x^2-2)}},#

#=1/2{(-x^2+16x-2)/{(x-8)(x^2-2)}}.#

# rArr dy/dx=-(y(x^2-16x+2))/{2(x-8)(x^2-2)}.#

But, #y={(x-8)/(x^2-2)}^(1/2).#

#:. dy/dx=-{(x-8)/(x^2-2)}^(1/2)*(x^2-16x+2)/{2(x-8)(x^2-2)}, i.e., #

# dy/dx=F'(x)=-((x^2-16x+2))/{2(x-8)^(1/2)(x^2-2)^(3/2)}.#

Enjoy Mayhs.!