The range of - 1 power of y = x?

The range of - 1 power of y = x?

y=x^-1=1/x.
The domain is defined as
(-∞,0)U(0,+∞)
Then the inverse function derivation, xy = 1, x = 1 / y, y = 1 / X
So the value field is equal to the definition field

Find the value range of the power X of y = 1-A to the power of 1 + a

Let y = f (x), it is easy to know that the definition domain of F (x) is R. f (x) = (1-A · x) / (1 + a · x) = - (a · x + 1-2) / (a · x + 1) - 1. Because a ·x {(0, + ··), it is easy to know that the value domain of F (x) is (- 1,1)!

Value range: y = 4 to the negative x power - 2 to the negative x power + 1

y=4^(-x) -2^(-x) +1
= [2^(-x) -1/2]^2 + 3/4
min y = 3/4
Range = [3 / 4, + ∞)

Xy = e to the (x + y) power to find the derivative of Y I don't understand. I don't know how e ^ (x + y) you are deriving, but we only learned to use logarithms to find such a derivation

(xy)'=(e^（x+y）'
y+xy'=e^(x+y)*(1+y')
y'=[e^(x+y)-y]/[1-e^(x+y)]

Find the derivative of implicit function 1, the third power of X + the second power of xy = y, 2, xcosy = LNY

The third power of 1. X + the second power of xy = y
Derivation 3x ^ 2 + y + X (dy / DX) = 2Y (dy / DX)
dy/dx=(3x^2+y)/(2y-x)
2.xcosy=lny
Derivation cosy xsiny * y '= y' / Y
y'=cosy*y/(1+xysiny)

How to find the derivative of e to the power of XY How to find the derivative of this formula? It's just a part of an equation Can you write down the process?

The derivative of X is y * e ^ (XY)
The derivative of Y is x * e ^ (XY)
The partial derivative of X, y is e ^ (XY) + XY * e ^ (XY)

What is the derivative of y = x to the power of X As above Can't fight the power It's the derivative of X in the upper right corner of X

Derivation of (x ^ x)
=Derivation of (e ^ (xlnx))
=[e^(xlnx)]*(lnx+x/x)
=(x^x)*(1+lnx)

Y = x to the power of X to find the derivative of X?

A:
Let z = x ^ x, then lnz = xlnx
(lnz)'=(xlnx)'
z'/z=lnx+1
So Z '= (x ^ x)' = (LNX + 1) * x ^ x
If y = x ^ (x ^ x), then LNY = (x ^ x) LNX
(lny)'=[(x^x)lnx]'
y'/y=(x^x)'lnx+(x^x)/x=(lnx+1)(lnx)*x^x+x^(x-1)
So y '= x ^ (x ^ x + x) * [(LNX) ^ 2 + LNX + 1 / x]

Y = SIN3 power X

∵sin3x
=sin(x+2x)
=sin2xcosx+cos2xsinx
=2sinx(1-sin²x)+(1-2sin²x)sinx
=3sinx-4sin^3x
ν y = SIN3 power x = (3sinx sin3x) / 4

The derivative y = 2 to the x power of e

y= 2^x * e^x
First you have to know the derivative of 2 ^ x, D / DX (2 ^ x) = ln (2) e ^ X
So,
dy/dx = ln(2)e^x * (e^x) + e^x * 2^x
= ln(2) e^2x + e^x * 2^x