How to calculate the derivative function of y = 1 / 3x ^ 3-2

How to calculate the derivative function of y = 1 / 3x ^ 3-2

y'=3*1/3x^(3-1)-0
=x^2

If the analytic function f (z) = u (x, y) + iy / (x ^ 2 + y ^ 2), then f '(1)=
Complex function problems (I need to solve the process) thank you

Finding the analytic region of the function f (z) = | x | + IY

u(x,y)=|x|,v(x,y)=y;
du/dx=-1 (x=0) ,du/dy=0;
dv/dx=0 ,dv/dy=1;
According to the Cauchy Riemann condition, the function is analytic in the complex plane where x > 0

Analysis of F (z) = x ^ 2-iy complex variable function

F (z) = u (x, y) + IV (x, y), now u = u (x, y) = x & # 178;, v = V (x, y) = - y, if we obtain partial derivatives for u and V respectively, then u / &; X = 2x, &; U / &; y = 0, &; V / &; X = = 0, &; V / &; y = - 1, so the partial derivatives of x = - 1 / 2 function are continuous and satisfy the Cauchy Riemannian bar

Verify that the following functions are harmonic functions, and find the analytic function w = f (z) = u + IV with Z = x + iy as the independent variable
u=(x-y)(x^2+4xy+y^2)
（1-i)z^3+ci;

After that meeting

Given the equation sin (XY) + X + y = 1, we determine the function y = y (x) and find y '

The derivation of the two sides is as follows
cos(xy)*(y+xy')+1+y'=0
y'[xcos(xy)+1]=-ycos(xy)-1
So, y '= - [ycos (XY) + 1] / [xcos (XY) + 1]

The set of all discontinuities of Z = sin (XY) \ \ (1-x2-y2) is
What's the format of the answer,

According to the definition field of function Z, all the discontinuities of Z = sin (XY) / (1-x & # 178; - Y & # 178;) are 1-x & # 178; - Y & # 178; = 0. These discontinuities are located on the unit circle, which should be expressed in the form of set: {(x, y) | X & # 178; + Y & # 178; = 1};

The function f (x) defined on R satisfies that f (x + y) = f (x) + F (y) is an odd function for any x, y ∈ R

It is proved that: let x = y = 0, substitute f (x + y) = f (x) + F (y), then f (0 + 0) = f (0) + F (0), that is, f (0) = 0. Let y = - x, substitute f (x + y) = f (x) + F (y), then f (x-x) = f (x) + F (- x), and f (0) = 0, then 0 = f (x) + F (- x). That is, f (- x) = - f (x) holds for any x ∈ R, so f (x) is an odd function

It is known that the function f (x) defined on (0, positive infinity) belongs to (0, positive infinity) for any x, Y. f (XY) = f (x) + F (y) is constant if 0

1. Obviously, f (1) = 0
For any x > 1, there is 0

It is known that the function f (x) defined on (0, positive infinity) belongs to (0, positive infinity) for any x, y, and f (XY) = f (x) + F (y), and when x > 1, f (x) < 0; f (2) = - 1. (1) prove that f (x) is a decreasing function on (0, positive infinity); (2) solve the inequality f (x) + F (x-3) > - 2

(1) Order 0