Question

How do you differentiate `((e^x)+x)^(1/x)`?

Answer 1

`d/dx (e^x+x)^(1/x)= (e^x+x)^(1/x){ (e^x+1)/(x(e^x+x)) -(ln(e^x+x))/x^2 }`

Explanation:

The best approach here is to use logarithmic differentiation to remove the variable exponent, as follows:

Let `y = (e^x+x)^(1/x)`

`=> ln \ y = ln \ {(e^x+x)^(1/x)}`
`" "= 1/x \ ln(e^x+x)`

We can now differentiate the LHS implicitly, and the RHS using the product rule to get:

`\ \ \ \ \ 1/y \ dy/dx = (1/x)(d/dx ln(e^x+x)) +(d/dx 1/x)(ln(e^x+x))`
`:. 1/y \ dy/dx = (1/x)( 1/(e^x+x) * (e^x+1))) +(-1/x^2)(ln(e^x+x))`
`:. 1/y \ dy/dx = (e^x+1)/(x(e^x+x)) -(ln(e^x+x))/x^2`
`:. dy/dx = (e^x+x)^(1/x){ (e^x+1)/(x(e^x+x)) -(ln(e^x+x))/x^2 }`