How do you differentiate y =(sqrtx-3)^3 using the chain rule?

1 Answer

dy/dx=3/(2sqrtx) xx(sqrtx-3)^2

Explanation:

y=(sqrtx-3)^3

Let
t=sqrtx-3

Let
u=sqrtx
(du)/(dx)=1/(2sqrtx)
v=3
(dv)/(dx)=0
t=u-v

dt/dx=(du)/(dx)-(dv)/(dx)

dt/dx=1/(2sqrtx)-0

dt/dx=1/(2sqrtx)

y=t^3

dy/dx=3t^2dt/dx

Substituting for t and dt/dx

dy/dx=3xx(sqrtx-3)^2 xx 1/(2sqrtx)

dy/dx=3/(2sqrtx) xx(sqrtx-3)^2