0
xy-e^x+e^y=1
xy-1=e^x-e^y
y+xy'=e^x-y'e^y
y'=(e^x-y)/(x+e^y)
Y power of E + xy = e to find the second derivative
Second order really troublesome, easy to make mistakes, had to ask in detail
When we find the bivalent y ', the Y' can be replaced by 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]
Find the derivative of the implicit function siny + e to the x power - XY power 2 = 0 Hope to be a little more detailed!
The derivative of an implicit function is to find the differential between the left and the right, and add a D,
Then write how many DX + how many dy = 0,
It would be nice to shift the term into the form of dy / DX = how much
Calculate the derivative y of the implicit function xy = e x-power-e y-th power,
xy=e^x-e^y
The derivation of both sides is as follows:
y+xy'=e^x-y'*e^y
The solution is as follows:
y'=(e^x-y)/(e^y+x)
It is known that y = y (x) is an implicit function determined by the equation xy = 1-e to the y-th power
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y=a^x
Y '= a ^ x * LNA. (a > 0, a is not equal to 1)
The x-th derivative of a F '(x power of a) = x power of a * ina How to push it
X power of a = ln of E = x times LNA of e
Using the derivative rule of composite function,
The derivative of a to the x power = e's [x times LNA] and then LNA = x power * LNA
Find the derivative of X to the power of X
Let y = x ^ x take the logarithm of both sides LNY = LNX ^ x = xlnx
(1 / y) * y '= (xlnx)' = 1 + LNX, so y '= y (1 + LNX) = x ^ x (1 + LNX)
How to find the derivative of X to the power of X As in the title,
If the original formula is y = x ^ x, we need to take the logarithm of both sides to become ln y = LNX ^ x = xlnx