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

1 Answer

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#