The solution to this is:
(xx)'=
(elnx to the power of x)'=
(exlnx)'=
(xlnx)'*exlnx=
[x(1/x) + 1(lnx)]*exlnx =
(lnx+1)*exlnx=
(lnx+1)*xx
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)
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.
Your expression simplifies to just x^2 {with the restriction that x > 0}. The derivative of x^2 is 2*x
If the function is (ln x)2, then the chain rules gives us the derivative 2ln(x)/x, with the x in the denominator. If the function is ln (x2), then the chain rule gives us the derivative 2/x.