Question

How do you differentiate `f(x)=ln(1/sqrt(xe^x-x))` using the chain rule.?

Answer 1

Using two power rules, we can simplify the expression and then proceed to derivate it more simply.

Explanation:

Two power rules to be remembered:

• `a^(-n)=1/(a^n)`
• `a^(m/n)=root(n)(a^m)`

Rewriting the expression: `f(x) = (xe^x-x)^(-1/2)`, then.

Using the chain rule, which states that `(dy)/(dx)=(dy)/(du)(du)/(dx)`, let's rename `u=xe^x-x` and derivate it:

`(df(x))/dx=-(1/2)u^(-3/2)(e^x+xe^x-1)=-(e^x+xe^x-1)/(2(xe^x-x)^(3/2))`

Just organizing:

`(df(x))/(dx)=(1-e^x(1+x))/(2(x(e^x-1))^(3/2))`