How do you solve: d/dx ((1+ 1/x)^x)

So you need to differentiate function `y(x)=(1+1/x)^x.` The problem is that both base and exponent contain variable `x.` To solve that we first need to take natural logarithm of both sides. ` `

`lny=ln((1+1/x)^x)`

Now we use the following rule for logarithms

`log_a b^c=c cdot log_a b`

Hence we get

`lny=xln(1+1/x)`

Now we can differentiate both sides, but we must keep in mind that `y` is a function of `x` .

`1/ycdoty'=ln(1+1/x)+x/(1+1/x)cdot(-1)/x^2`

`(y')/y=ln(1+1/x)-1/(x+1)`

Now we multiply whole expression by `y=(1+1/x)^x.`