how would you find the derivative of y = x^(1-x) ?

Since you're starting with the equation y = x^(1-x), Your best bet is logarithmic differentiation. Logarithmic differentiation basically asks you to first take the natural log of both sides and then derive both sides as a way of obtaining the derivative. So let's first establish this.

y = x^(1-x) -> ln |y| = ln |(x^(1-x))|

Now, by the properties of natural log, an exponent of a function is the same as multiplying its natural log, like so:

ln |(x^(1-x))| = (1-x)*ln |x| = ln|x| - xln|x|

Next, we derive both sides. Since y is still written as ln|y|, we will have to derive it implicitly. Remember that the derivative of the natural log is the same as the reciprocal of the number inside ((ln|x|)' = 1/x). This means that the derivative of ln|y| = y'/y by the chain rule.

y'/y = (ln|x| - xln|x|)' = 1/x - (1 + ln|x|) = 1/x - 1 - ln|x| (The derivative of x*ln|x| is going to require the product rule)

Lastly, we multiply both sides by y to get y' by itself.

y' = (1/x - 1 - ln|x|)*(x^(1-x))

And you're done! You could distribute the x^(1-x) but most professors won't require it since you could plug in x and solve it at this point. Hope this helps!

ln y=(1-x)lnx

1/y*y'=-lnx+(1-x)/x

y'=y(-lnx+(1-x)/x)

y'=x^(1-x)(-lnx+(1-x)/x)