Let the function y = y (x) be determined by the equation E ^ y + xy = e, and find y '(0)

Let the function y = y (x) be determined by the equation E ^ y + xy = e, and find y '(0)

The derivative of X on both sides is y '* e ^ y + y + XY' = 0,
Let x = 0 in the original equation, then y = 1,
Therefore, y '+ 1 = 0 can be obtained by substituting x = 0 and y = 1 into the above formula,
That is y '(0) = - 1

Let y = y (x) be determined by the equation E ^ XY + y ^ 3-5X = 0, and find y '(0)
Is e ^ XY derived from ye ^ XY or (y + XY ') e ^ XY

The latter is right
(e^xy)'=e^xy*(xy)'
(xy)'=(x)'y+xy'=y+xy'
Immediate solution

Let the function y = y (x) be determined by the equation E * y + xy = e, and find y '(0)

Find the derivative of X for the equation
(e*y)'+(xy)'=(e)'
ey'+(y+xy')=0
y'=-y/(e+x)
When x = 0, y = 1, y '= - Y / (E + x) = - 1 / (E + 0) = - 1 / E

Let e ^ y + xy = e determine the function y = y (x) and find y '| x = 0

e^y+xy=e
Derivation of X
e^y*y'+(y+xy')=0
(e^y+x)y'=-y
y'=-y/(e^y+x)
When x = 0, y = 1
y'=-1/(e+0)=-1/e

Let X-Y-Z = 19, X2 + Y2 + Z2 = 19, then yz-zx-xy=______ ．

If X-Y-Z = 19 is squared, then: (X-Y-Z) 2 = 361, that is, X2 + Y2 + z2-2xy-2xz + 2yz = 361, ∵ x2 + Y2 + Z2 = 19, ∵ x2 + Y2 + z2-2xy-2xz + 2yz = 19 + 2 (YZ XY XZ) = 361, then YZ XY XZ = 361 − 192 = 171

What is the relationship among the existence of partial derivatives, differentiability and continuity of functions?

In the case of one variable, differentiable = differentiable → continuous, differentiable must be continuous, otherwise not necessarily. Binary is not satisfied. In the case of binary, partial derivative exists and continuous, function differentiable, function continuous; partial derivative does not exist, function non differentiable, function not necessarily continuous. Function differentiable, partial derivative exists, function continuous; function non differentiable, partial derivative does not necessarily exist, Function is not necessarily continuous. If function is continuous, partial derivative may not exist and function may not be differentiable. If function is discontinuous, partial derivative may not exist and function may not be differentiable

Why the existence and continuity of partial derivative of multivariate function at a point can not prove that the function is differentiable at that point?
It's the one in example 10.8 of science and engineering p254 of Chen Wendeng's 2010 postgraduate entrance examination mathematics review guide. I feel that C and D are the same. Why is D right? Are there many questions in Wendeng's book?

Differentiability is more strict than differentiability. Differentiability is for a certain independent variable, while differentiability is for all independent variables. Multivariate function has multiple independent variables. To be differentiable, function must be differentiable for all independent variables at the change point. From the perspective of image, differentiability is from one direction, while differentiability is from multiple directions

The relationship between the differentiability of partial derivative and directional derivative of multivariate function
If we are differentiable, then the partial derivative exists, and so on

The existence of partial derivatives is not necessarily differentiable, but differentiable partial derivatives must exist
Only if the partial derivative exists and is continuous, it must be differentiable

Multivariate function is differentiable, why can it be differentiable by adding partial derivative?

Differentiable, but it may break at some point. It must be a complete function, that is, continuous to be differentiable

I want to know the relationship between differentiable, integrable, continuous partial derivative, continuous function and differentiable partial derivative. Note that this is in partial derivative

First of all, partial derivatives are not necessarily continuous or differentiable. Continuous functions are not necessarily partial derivatives or differentiable. If they are differentiable, partial derivatives must be obtained. If they are continuous and partial derivatives can be obtained, functions are continuous and differentiable. Continuous functions on closed areas with areas are integrable, etc