U = x (Z + y) z = sin (x + y) finding the second order partial derivative σ 2U / σ x σ y

U = x (Z + y) z = sin (x + y) finding the second order partial derivative σ 2U / σ x σ y

σu/σx=(z+y)+x(σz/σx+0)=z+y+xcos(x+y)
σ2u/σxσy=σz/σy+1-xsin(x+y)=cos(x+y)+1-xsin(x+y)

Let u = f (lnxy, sin (XY)), find the partial derivatives of X and y with respect to u

du/dx=f1/x+y*cos(xy)*f2
du/dy=f1/y+x*cos(xy)*f2

Find the partial derivatives of the following functions: z = sin (x under y radical)

のz/のx=cos(y√x)·[y/(2√x)]
=[y/(2√x)]cos(y√x)
のz/のy=cos(y√x)·√x
=√xcos(y√x)

Given that the derivative of function f (x) is f '(x) = 3x ^ 2-6x, and f (0) = 4, find the expression of function f (x)

The derivative function of ∵ y = f (x) is f '(x) = 3x ^ 2-6x ∵ f (x) = x ^ 3-3x ^ 2 C (C is a constant) and ∵ f (0) = 4 ∵ C = 4 ∵ f (x) = x ^ 3-3x ^ 2 4. Let f' (x) < 0, the monotone decreasing interval of 0 < x < 2 ∵ f (x) is (0,2), monotone increasing interval is (- ∞, 0), [2, ∞) and f (0) = 4 > 0, f (2

Finding the partial derivative of binary function (2x + y) ^ (2x + y) to X
555555 is a variable. How to say it? Is it better to split it into (2x + y) ^ (2x) * (2x + y) ^ y first,

Let z = (2x + y) ^ (2x + y)
lnz=(2x+y)ln(2x+y)
Finding partial derivatives of X
We get Z '/ z = 2ln (2x + y) + 2
So partial Z / partial x = [2ln (2x + y) + 2] (2x + y) ^ (2x + y)

On the existence and continuity of differentiable, differentiable and partial derivatives
For example: differentiability is not necessarily continuous, the existence of partial derivatives is not necessarily differentiable
Hope to give all the connections between them~

Partial derivatives are continuous = > differentiable {= > partial derivatives exist 1
=>Function continuous 2
It doesn't matter between 1 and 2

Limit, continuity, existence of partial derivative, partial derivative, differentiability

The partial derivatives FX, FY are continuous at point (x0, Y0) (1) z = f (x, y) differentiable at point (x0, Y0) and DZ = ADX + bdy (2) f (x, y) continuous at point (x0, Y0) (3) z = f (x, y) differentiable at point (x0, Y0), and FX = a, FY = B (4) 1 can push 2,2 can push 3,2 can push 42 can't push 1,3 can't push 2,3 can't push 4 can't

The difference between the existence of partial derivative and the continuity of partial derivative
If the left and right limits of the partial derivatives at one point are equal, then the partial derivatives exist. How can we prove that the partial derivatives are continuous? Is the partial derivatives at that point equal to the left and right limits?

This is actually a continuous problem of proof
If the left and right limits are equal, the partial derivative exists. But the limit at this time is not necessarily equal to the derivative value of the point, OK?
To prove that the partial derivative is continuous is to prove that the left and right limits are equal and equal to the partial derivative value of the point
That is to say: the partial derivative at that point is equal to the left and right limit

How to judge the existence of partial derivative
What are the sufficient and necessary conditions for the existence of partial derivatives of multivariate functions?

The necessary and sufficient condition for the existence of partial derivatives of multivariate functions at x0 is
The existence of (t tends to 0) Lim [f (x0 + T) - f (x0)] / T is the same for other independent variables
The relation between partial derivative and continuity of multivariate function is neither necessary nor sufficient

Finding the partial derivative of F (x, y) = XY (x ^ 2-y ^ 2) / (x ^ 2 + y ^ 2)

First, find the total derivative of the function as follows:
df(x,y)={[xy(x^2-y^2)]'(x^2+y^2)-xy(x^2-y^2)(x^2+y^2)'}/(x^2+y^2)^2
={[(xy)'(x^2-y^2)+(xy)(x^2-y^2)'](x^2+y^2)-xy(x^2-y^2)(2xdx+2ydy)}/(x^2+y^2)^2
={[(ydx+xdy)(x^2-y^2)+xy(2xdx-2ydy)](x^2+y^2)-2xy(x^2-y^2)(xdx+ydy)}/(x^2+y^2)^2
=【y(x^4-4x^2y^2-y^4)/(x^2+y^2)^2】dx-【x(x^4-5y^4)/(x^2+y^2)^2】dy
The former DX is the partial derivative to x, and the latter dy is the partial derivative to y