How to solve the implicit function problem?

This is the content of multivariate function differentiation, f'x is the partial derivative of f to x, and the same goes for f'y，f'x(0,1) is the rate of change of f near 0 when y = 1 (over a plane parallel to the plane xoz (0,2,0)), and the same goes for f'y(0,1), you bring in the fixed coordinates, and then find the derivative, the result is the same, but it is obvious that the main thing is that the position of the point of the derivative function you bring in is different, and the specific slope is intuitively expressed, you can use some commonly used mathematical software such as matlab to see, the partial derivative is best combined with the spatial coordinates.

f(x,y) denotes that the function is a binary function, where y and x are already equivalent variables, and y is regarded as a constant when the x is derived, and x is regarded as a constant when the y derivative is obtained.

But your question here is: I also added a sentence x + y -1 = 0, so doesn't it mean that f(x,y) is always equal to 0.

f(x,y)=x +y-1 is a binary function, f'x，f'y is the partial derivative of the function f(x,y).

Find the partial derivative f'x, take the y term as a constant, and get f'x=2x, the same goes for f'y=2y

When finding f(0,1), substitute x=0,y=1 into f(x,y)=x +y -1 to get f(0,1)=0+1-1=0

f‘y(0,1)=2*1=2

This is a condition for the existence of implicit functions.

The steps are as follows:1On both sides of the equation, first find the first-order partial derivative for x to get the first-order partial derivative of z about x, and then solve the first-order partial derivative of z about x.

2.Find the partial derivative for x on both sides of the equation where the first-order partial derivative was originally sought. This equation must contain both a first-order and a second-order partial derivative of x.

Finally, substituting the first-order partial derivative solved in 1 into it, we can get an equation that contains only the second-order partial derivative. Just solve it. <>

If z=f(y) and y=g(x), then z=f[g(x)], the result is an imitation composite function, and if the relationship between x and y is f(x,y)=0, that is, the function formula y equal to x cannot be directly obtained, then it is an implicit function.

For a case that has been determined to exist and is derivable, we can use the law of the chain of large clusters derived by composite functions.

The derivatives of x are taken on both the left and right sides of the equation, since y is actually a function of x.

So you can get it directly with y'and then simplify to get y'expression.

Fit within the proximity of a point of the original equation, under the premise that the function f(x,y) is continuously differentiable.

What kind of additional conditions can make the original equation determine a unique function y=(x), which is not only single-valued continuous, but also continuously differentiable.

Its derivative is determined by the full. The implicit function existence theorem is used to conclude that such a condition is not only necessary, but sufficient.

Derivative of x. 2yz'xe^(2yz)+1+z'x = 0 to get z'x= 1 (2ye (2yz)+1) is the derivative of y. 2z+2yz'y)e^(2yz)+2y+z'y=0 to get z'y= 2(y+ze (2yz)) 2ye (2yz)+1) and then substitute x=1 2 and y=1 2 into the formula.

E + 1 2 + 1 4 + z = 7 4 can be obtained, and z = 0 is z'x=－1/2，z'y=－1/2

Then set the book formula to get dz

First of all, if you just say x 2+y 2-1=0, this cannot be an implicit function.

It should be said that it is not a function at all, and conditions should be added to the problem, such as: y 0 and so on.

Under such conditions, since it can be determined that there is an implicit function, of course, the explicit function can be written.

y=(1-x^2)^(1/2)

Derivative: y'=-x/y

For x 2 + y 2-1 = 0

Derivative: 2x+2yy'=0

y'=-x/y

It can be seen that the two are the same.

But be careful: it's all determined by the added condition, if it's not y 0 but something else.

Then, the explicit function of the dissolution should be adjusted accordingly.

If you don't understand, please ask.

You don't strictly put it, only the upper part, the lower half is not included, in fact, it is okay, but the calculation or expression is not so concise, if this is expressed and calculated in the form of implicit functions, it is obviously much more concise.

Definitely not, because your method only gives the positive part of the function y, in fact, y can be negative....

No way! y= (1-x 2) has two curves.

1.Personally, I think that the meta should refer to the number of variables in the function, so f(x,y)=0 is binary, and f(x,y,z) is ternary. It's not a matter of nature, it's just a matter of habit and definition, and it's okay to define it the way you want it.

2.The proof is as follows.

Differentiation on both sides of f(x,y,z)=0, obtain.

f to x partial derivation) * dx + (f to y partial derivance) * dy+ (f to z partial derivation) * dz = 0 If (z to x partial derivation) then let dy=0 (partial derivative definition, y unchanged), i.e., (f to x partial derivation) * dx + (f to z partial derivation) * dz = 0 to obtain (z to x partial derivation) = dz dx|dy=0=-(f-x-bias) (f-z-bias)

The same is true for (z to y partial derivation).