Question

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

Answer 1

`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)`