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

Question

Emilia Aguilar · Accepted Answer

#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)#

Emma Aldridge · Answer

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

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.!

Elijah Abram · Answer

undefined To find the derivative of $ F(x) = \sqrt{\frac{x-8}{x^2-2}} $, you can use the quotient rule and the chain rule. The quotient rule states that if $ f(x) = \frac{g(x)}{h(x)} $, then $ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} $. The chain rule states that if $ g(x) $ and $ h(x) $ are differentiable functions, then $ \frac{d}{dx}[g(h(x))] = g'(h(x)) \cdot h'(x) $.

First, find the derivatives of $ g(x) = \sqrt{x-8} $ and $ h(x) = x^2-2 $. Then apply the quotient rule to find the derivative of $ F(x) $.

$ g'(x) = \frac{1}{2\sqrt{x-8}} $
$ h'(x) = 2x $

Now, apply the quotient rule:

$ F'(x) = \frac{\frac{1}{2\sqrt{x-8}}(x^2-2) - \sqrt{x-8}(2x)}{(x^2-2)^2} $

Simplify:

$ F'(x) = \frac{x^2 - 2}{2(x^2-2)^{\frac{3}{2}}} - \frac{2x\sqrt{x-8}}{(x^2-2)^2} $

This is the derivative of $ F(x) $.