How do you find the derivative of #G(x) = sqrtx (x^2 – x)^3#?

2 Answers
Nov 8, 2017

Answer:

Use chain rule and product rule. #=(x^2-x)^2((x^2-x)/(2sqrtx) + (6x-3)sqrtx)#

Explanation:

As written, we use both the product rule and chain rule.

product rule: #f(x) = g(x) * h(x) -> f'= g'*h + g*h'#
chain rule: #f(x) = g(h(x)) -> (df)/dx= (dh)/(dx)*(dg)/(dh)#

Then with #G(x) = sqrtx (x^2-x)^3...#

#(dG)/(dx) = d/dx (sqrt x)*(x^2-x)^3 sqrtx*d/dx ((x^2-x)^3) = 1/(2sqrtx) (x^2-x)^3 + sqrtx *(2x-1) *3(x^2-x)^2#

#=(x^2-x)^2((x^2-x)/(2sqrtx) + (6x-3)sqrtx)#

Nov 8, 2017

Answer:

#G'(x)=1/2sqrtx(x^2-x)^2(13x-7)#

Explanation:

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

#"given "G(x)=g(x)h(x)" then"#

#G'(x)=g(x)h'(x)+h(x)g'(x)larr"product rule"#

#g(x)=sqrtx=x^(1/2)rArrg'(x)=1/2x^(-1/2)larr"chain rule"#

#h(x)=(x^2-x)^3larr"differentiate using chain rule"#

#rArrh'(x)=3(x^2-x)^2 (2x-1)#

#G'(x)=3x^(1/2)(x^2-x)^2(2x-1)+1/2(x^2-x)^3x^ (-1/2)#

#color(white)(rArrG)=1/2x^(-1/2)(x^2-x)^2[6x(2x-1)+x^2-x]#

#color(white)(xxxx)=1/2x^(-1/2)(x^2-x)^2(13x^2-7x)#

#color(white)(xxxx)=1/2sqrtx(x^2-x)^2(13x-7)#