How do you differentiate #f(x)= 2x^3(x^3 - 3)^4 # using the chain rule?

1 Answer
Nov 19, 2015

Answer:

Use both the Product and the Chain Rules to get
#(df(x))/(dx) = (x^3-3)^3*(24x^5+6x^4-18x)#

Explanation:

Given #f(x) = 2x^3(x^3-3)^4#

Let #f(x) = p(x)*q(x)#
where #p(x)=2x^3# and #q(x)=(x^3-3)^4#

Further let #q(x) =r(s(x))#
where #r(x)=x^4# and #s(x)=x^3-3#

#(dp(x))/(dx) = (d 2x^3)/(dx) = 6x#

#(ds(x))/(dx) = (d x^3-3)/(dx)=3x^2#

#(dr(x))/(dx) = (d x^4)/(dx) = 4x^3#
#rArr (d r(s(x)))/(ds(x)) = 4(x^3-3)^3#

The Chain Rule tells us that
#(dq(x))/(dx) = (d r(s(x)))/(dx) = (d r(s(x)))/(d s(x))*(d s(x))/(dx)#
#color(white)("XXX")=4(x^3-3)^3*3x^2 = 12x^2(x^3-3)^3#

The Product Rule tells us that
#color(white)("XXX")(df(x))/(dx) =(d(p(x) * q(x)))/(dx) = q(x) * (dp(x))/(dx)+p(x) * (dq(x))/(dx)#

#color(white)("XXXXXX")=(x^3-3)^4*6x + 2x^3*12x^2(x^3-3)^3#

#color(white)("XXXXXX")=(x^3-3)^3*[(x^3-3)*6x + 2x^3*12x^2]#

#color(white)("XXXXXX")=(x^3-3)^3(24x^5+6x^4-18x)#