If #f(x)= sqrt(x-2 # and #g(x) = 1/x #, what is #f'(g(x)) #?

1 Answer
Jul 10, 2016

Answer:

#-1/(2x^2sqrt(1/x-2)#

Explanation:

Before differentiating we require to establish f(g(x))

Substitute x #=1/x"into f(x)"#

#rArrf(g(x))=f(1/x)=sqrt(1/x-2)#

now differentiate using the #color(blue)"chain rule combined with power rule"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(d/dx(f(g(x)))=f'(g(x))g'(x))color(white)(a/a)|)))........ (A)#

Express #sqrt(1/x-2)" as " (x^-1-2)^(1/2)#
#"--------------------------------------------------"#
#f(g(x))=(x^-1-2)^(1/2)rArrf'(g(x))=1/2(x^-1-2)^(-1/2)#

and #g(x)=x^-1-2rArrg'(x)=-1x^(-2)=-x^(-2)#
#"--------------------------------------------------------"#
Substitute these values into (A)

#rArrf'(g(x))=1/2(x^-1-2)^(-1/2).(-x^-2)#

Expressing the answer with positive indices

#rArrf'(g(x))=-1/(2x^2sqrt(1/x-2)#