Question

How do you use the chain rule to differentiate `root9(e^(6x))`?

Answer 1

`f(x)=2/3root(3)(e^(2x))`

Explanation:

According to chain rule when `f=f(g(h(x)))`,

`(df)/(dx)=(df)/(dg)xx(dg)/(dh)xx(dh)/(dx)`

Here `f(x)=root(9)g(x)`, `g(x)=e^(h(x))` and `h(x)=6x`

and `(df)/(dg)=1/9(g(x)^((1/9-1)))`

`(dg)/(dx)=e^(h(x))` and `(dh)/(dx)=6`

Hence as `f(x)=root(9)(e^(6x))`

`(df)/(dx)=1/9(g(x)^((1/9-1)))xxe^(h(x))xx6`

= `6/9(e^(6x))^((1/9-1))e^(6x)`

= `6/9e^(6x xx(1/9-1))e^(6x)`

= `6/9e^(6x xx(1/9-1)+6x)`

= `6/9e^((6x)/9)`

= `2/3root(3)(e^(2x))`