Question

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

Answer 1

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)`

Answer 2

`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)`