I get x*x^x-1 + lnx*x^x = x^x + x^xlnx = x^x * (1+lnx)
Here, ^ is power; * = times; ln = natural logratithm ( base e)
Chat with our AI personalities
x^0 = 1 for all x. The derivative of 1 is always zero.
e^(-2x) * -2 The derivative of e^F(x) is e^F(x) times the derivative of F(x)
Your expression simplifies to just x^2 {with the restriction that x > 0}. The derivative of x^2 is 2*x
the derivative of 1x would be 1 the derivative of x to the power of 1 would be 1. the derivative of x+1 would be 1 the derivative of x-1 would be 1 im not sure what you are asking, but however you put it, it's 1.
y = e^ln x using the fact that e to the ln x is just x, and the derivative of x is 1: y = x y' = 1