Question

What is the derivative of `(x+1)^x`?

Answer 1

`(x+1)^x(x/(x+1)+lnx)`

Explanation:

`y=(x+1)^x`

lets use `ln` in the two sides of the equation,

`lny=xln(x+1)`

now lets differentiate the two sides in respect of `x`

`1/y(dy)/(dx)=xd/(dx)(ln(x+1))+lnxd/(dx)(x)`

`or,1/yd/dx=x*1/(x+1)d/dx(x+1)+lnx`

`or,1/y(dy)/(dx)=x/(x+1)+lnx`

`or,(dy)/(dx)=y(x/(x+1)+lnx)`

`or,(dy)/(dx)=(x+1)^x(x/(x+1)+lnx)`

Answer 2

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

Explanation:

I think the easiest way to do these kinds of problems (with a variable power) without memorizing the formula is through implicit differentiation.

`y=(x+1)^x`

Take the natural logarithm of both sides.

`ln(y)=ln((x+1)^x)`

Rewrite using logarithm rules.

`ln(y)=xln(x+1)`

Differentiate both sides with respect to `x`.

The left side will spit out a `dy/dx` term, thanks to the chain rule. The right side will require the product rule.

`1/y(dy/dx)=ln(x+1)d/dx(x)+xd/dx(ln(x+1))`

Note that differentiating `ln(x+1)` requires chain rule as well, but since `d/dx(x+1)=1` it doesn't make a visible difference.

`1/y(dy/dx)=ln(x+1)+x(1/(x+1))d/dx(x+1)`

`1/y(dy/dx)=ln(x+1)+x/(x+1)`

Now, solve for `dy/dx` by multiplying both sides by `y`. However, remember that `y=(x+1)^x`.

`dy/dx=(x+1)^x(ln(x+1)+x/(x+1))`