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.
It is a physics that can be simplified. The simplified version is y = e^(2lnx) lny = ln(e^(2lnx)) lny = 2lnx * lne lny = 2lnx * 1 lny = ln(x^2) y = x^2 x ^ 2 = e^(2lnx).
The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x
The first derivative of e to the x power is e to the power of x.
The derivative of ex is ex
e^(-2x) * -2 The derivative of e^F(x) is e^F(x) times the derivative of F(x)
Taking the derivative: 2lnx+2, 2(lnx+1). Extrema occur at y'=0. 2(lnx+1)=0,lnx=-1 x=e^(-1)=1/e y=-2/e So the extrema occurs at (1/e,-2/e)
I assume you mean 27 times e to the power x. 1) You take out the constant out. So, the derivative is 27 times the derivative of (e to the power x).2) You use the rule for the exponential function.
d/dx (e-x) = -e-x
e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
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
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