Y / x = ln (XY) find dy / DX (xy-y^2)/(xy+x^2)

Y / x = ln (XY) find dy / DX (xy-y^2)/(xy+x^2)

Two side derivation
(y'x-y)/x^2=(y+xy')/xy
xy+x^2y'=xyy'+y^2
y'=(xy-y^2)/(xy+x^2)

What is the value range of the function y = the 4th power of SiNx + the 4th power of cosx?

y=[(sinx)^2+(cosx)^2]^2-2(sinx)^2+(cosx)^2
=1^2-1/2*(2sinxcosx)^2
=-1/2*(sin2x)^2+1
=-1/2*(1-cos4x)/2+1
=1/2*cos4x+1/2
-1

Find the function y = f (x) = (1 / 4) to the power of X - (1 / 2) to the power of X + 1, and X belongs to the value range of [- 3,2]

Let t = (1 / 2) to the power of X (1 / 4)

Given the function f (x) = the x power of 10 + the x power of 10 / 10 to the - x power of - 10, find the value range

It is easy to know that the definition field of function f (x) = [10 ^ X-10 ^ (- x)] / [10 ^ x + 10 ^ (- x)]. Is R. it can be assumed that t = 10 ^ x.x ∈ R, ∵ T > 0. Therefore, the problem can be reduced to finding the value field of function g (T) = [T - (1 / T)] / [T + (1 / T)] on (0, + ∞). ? g (T) = (T) ²- 1)/(t ²+ 1),∴g(t)-1=-2/(t ²+ 1).∵t＞0.===>t ²+ 1＞1.===>0＜1/(t ²+ 1)＜1.===>0＜2/(t ²+ 1) < 2. = = > 0 < 1-g (T) < 2. = = > - 1 < g (T) < 1. ‡ the value range of the original function f (x) is (- 1,1)

The value range of function f (x) = [3 (x power) - 1] / [3 (x power) + 1] is

Domain x ∈ R
F (x) = [3 (x power) - 1] / [3 (x power) + 1]
F (x) = 9 (x power) - 1
Let u = 9 (x power)
The value range of u is (0, + ∞)
The value range of F (x) = 9 (x power) - 1 is (1, + ∞)

What is the derivative of cosx under the third root?

[(cosx)^(1/3)]'=(1/3)*[(cosx)^(-2/3)]*(-sinx)