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Find the second derivative of e to the power of Y + xy = E
Seeking second order is really troublesome and easy to make mistakes, so I have to ask in detail
When finding bivalent y ', y' can be replaced by the answer of the first derivative
Total derivative: let z = arctan (XY), and y = e * x power, find DZ / DX Let z = arctan (XY), and y = e * x power, find DZ / DX,
That is, z = arctan (Xe ^ x)
dz/dx={1/[1+(xe^x) ²]}* (xe^x)'
=(e^x+xe^x)/[1+(xe^x) ²]
The inverse function of the function y = INX · sin2x (x > 0) is
1/x*sin2x+2lnx*cos2x
Why is the derivative of y = INX 1 / X and how
Hello, everyone. I want to ask you, why is the derivative of y = ln x 1 / x? How did it come about?
Y=ln X
X=e^Y
dX=e^YdY
dY/dX=1/e^Y=1/X
So: y '= 1 / X
The maximum value of function y (x) = INX / x-x needs to be calculated in detail
The definition field is x > 0
Y '= (1-lnx) / x ^ 2-1 = (1-lnx-x ^ 2) / x ^ 2 = 0, G (x) = 1-lnx-x ^ 2 = 0,
Because g (x) is a monotone decreasing function and G (1) = 0, f (x) has only one pole x = 1
At x1, f (x) decreases
Therefore, f (1) = - 1 is the maximum value and also the maximum value
The maximum value of the function f (x) = INX / X is
Derivative
Get: (1-lnx) / X ²= 0
Because LNX exists, x > 0
x=e
Maximum value of F (x) = 1 / E
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