Question

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

Answer 1

Chain rule states that if `f(g(h(x)))`,

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

Hence, here for `f(x)=ln(1/(sqrt(e^x-x)))`

`(df)/(dx)=`

`1/(1/(sqrt(e^x-x)))xx(-1/((sqrt(e^x-x))^2))xx1/2(e^x-x)^(-1/2)xx(e^x-1)`

`=sqrt(e^x-x)xx(-1)/((e^x-x))xx1/(2sqrt(e^x-x))xx(e^x-1)`

= `(1-e^x)/(2(e^x-x))`