Your expression simplifies to just x^2 {with the restriction that x > 0}. The derivative of x^2 is 2*x
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The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x
e^(-2x) * -2 The derivative of e^F(x) is e^F(x) times the derivative of F(x)
d/dx (e-x) = -e-x
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
2.71828183 ==So the derivative of a constant is zero.If you have e^x, the derivative is e^x.