Question

How do you differentiate `f(x)=sqrt((ln(-x+3))^3` using the chain rule.?

Answer 1

`(df)/(dx)=-(3(ln(3-x))^2)/(2(3-x)sqrt((ln(3-x))^3)`

Explanation:

Chain Rule - In order to differentiate a function of a function, say `y, =f(g(x))`, where we have to find `(dy)/(dx)`, we need to do (a) substitute `u=g(x)`, which gives us `y=f(u)`. Then we need to use a formula called Chain Rule, which states that `(dy)/(dx)=(dy)/(du)xx(du)/(dx)`. In fact if we have something like `y=f(g(h(k(x))))`, we can have `(dy)/(dx)=(dy)/(df)xx(df)/(dg)xx(dg)/(dh)xx(dk)/(dx)`

Here we have `f(x)=sqrt((ln(-x+3))^3)` or say

`f(x)=sqrt(g(x))`, where `g(x)=(h(x))^3`, `h(x)=ln(k(x))` and `k(x)=-x+3`

and `(df)/(dg)=1/(2sqrt(g(x))`, `(dg)/(dh)=3(h(x)^2)`, `(dh)/(dk)=1/(k(x))` and `(dk)/(dx)=-1`

hence `(df)/(dx)=1/(2sqrt(g(x))xx3(h(x)^2)xx1/(k(x))xx(-1)`

= `1/(2sqrt((ln(-x+3))^3))xx3(ln(-x+3))^2xx1/((-x+3))xx(-1)`

= `-(3(ln(-x+3))^2)/(2(-x+3)sqrt((ln(-x+3))^3)`

= `-(3(ln(3-x))^2)/(2(3-x)sqrt((ln(3-x))^3)`