How do you differentiate # y =sqrt((3x-9)^3 # using the chain rule?

1 Answer
Jan 8, 2016

Answer:

#h'(x)=d/dxsqrt((3x-9)^3)=9/2sqrt(3x-9)#

Explanation:

given

#h(x)=f(g(x))=(f@g)#

The chain rule is

#h'(x)=(f@g)'=f'(g(x))*g'(x)#

#h(x)=y=sqrt((3x-9)^3)=((3x-9)^3)^(1/2)=(3x-9)^(3/2)#

then
#f(x)=( )^(3/2)#

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

#g(x)=(3x-9)#

#g'(x)=3#

Apllying the rule

#h'(x)=3/2(3x-9)^(3/2-1)*3=9/2(3x-9)^((3-2)/2)=9/2(3x-9)^(1/2)=#
#9/2sqrt(3x-9)#

Alternatively:

#f(x)=sqrt()#

#f'(x)=1/(2*sqrt())#

#g(x)=(3x-9)^3#

it is a composite function too

#g'(x)=3(3x-9)^(3-1)*d/dx(3x-9)=3(3x-9)^2*3=#
#=9(3x-9)^2#

then:

#h'(x)=(1/(2*sqrt((3x-9)^3)))9*(3x-9)^2=#
#=9/2*((3x-9)^2/(sqrt((3x-9)^2*(3x-9))))=#
#=9/2*((3x-9)^color(blue)cancel(2)/(color(blue)cancel((3x-9))*sqrt(3x-9)))=#
#=9/2(3x-9)/sqrt(3x-9)=9/2sqrt((3x-9)^color(blue)cancel(2)/color(blue)cancel((3x-9)))=#
#=9/2sqrt(3x-9)#