How do you differentiate #f(x)=(4x-x^2)^(1/2)/x^2# using the chain rule?

1 Answer
Dec 14, 2016

Answer:

#f'(x)=(x-6)/(x^2(4x-x^2)^(1/2)#

Explanation:

We will have to use the chain rule but before that we have to use the #color(blue)"quotient rule"#

#color(orange)"Reminder " "If " f(x)=(g(x))/(h(x))" then "#

#color(red)(bar(ul(|color(white)color(black)(f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2)|)))larr" quotient rule"#

#color(orange)"Reminder " " If " f(x)=g(h(x))" then"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(f'(x)=g'(h(x)).h'(x))color(white)(2/2)|)))larr" chain rule"#

differentiate using the #color(blue)"quotient rule"#

#g(x)=(4x-x^2)^(1/2)" and using the chain rule"#

#g'(x)=1/2(4x-x^2)^(-1/2).d/dx(4x-x^2)#

#=1/2(4x-x^2)^(-1/2)(4-2x)=1/2(4x-x^2)^(-1/2).2(2-x)#

#=(2-x)(4x-x^2)^(-1/2)#

#" and " h(x)=x^2rArrh'(x)=2x#
#"-------------------------------------------------------------"#

#rArrf'(x)=(x^2(2-x)(4x-x^2)^(-1/2)-(4x-x^2)^(1/2).2x)/x^4#

simplifying the numerator.

#=(x(4x-x^2)^(-1/2)[x(2-x)-(4x-x^2).2))/x^4#

#=(x(4x-x^2)^(-1/2)[2x-x^2-8x+2x^2])/x^4#

#=(x(4x-x^2)^(-1/2)(x^2-6x))/x^4=(cancel(x^2)(4x-x^2)^(-1/2)(x-6))/(cancel(x^4)x^2)#

#=(x-6)/(x^2(4x-x^2)^(1/2))#