If y=xxy = x^xy=xx for x>0x > 0x>0, what is the derivative dydx\frac{dy}{dx}dxdy?
xxln(x)x^x \ln(x)xxln(x)
xx(1+ln(x))x^x(1 + \ln(x))xx(1+ln(x))
x⋅xx−1x \cdot x^{x-1}x⋅xx−1
ln(x)+1\ln(x) + 1ln(x)+1