The answer to the question of finding the implicit function in high numbers has been given, and the

Not multiplying xy, because z is a function of xy, and xz as a whole takes the derivative of x. Find the x-derivative step by step, and finally sort out the result.

Derivative of y: on both sides of the system of equations:

2y + z+y z'y - z'y t^2 - 2zt t'y = 0 ==> (y-t^2) z'y - 2zt t'y = -2y-z

t'y e^z + t e^z z'y + z'y sint + z cost t'y = 0 ==> (t e^z + sint) z'y + e^z + z cost) t'y = 0

Solution: z'y = -(2y+z)(ez + zcost)/w; z't = (2y+z)(t e^z + sint)/w

u'y = f'y + f't t'y + f'z z'Y Answer: C

xy＋xz＋yz=0

The left and right sides are derived from x.

y＋z＋x∂z/∂x＋y∂z/∂x=0

z/∂x=－(y＋z)/(x＋y)

When x=0 and y=1, z=0

So z x|(x=0，y=1) =－1/1=－1

Summary. Both sides simultaneously derive y+xy for x'-eˣ+eʸy'＝0y'Y dx (e -y) (x+e) When x 0, y 0 is brought into the above equation dy dx 1

Send a picture? Hello dear customers, there are three questions in one round of middle and high school mathematics and two questions in one round of advanced mathematics, thank you for your understanding!

Fourth, answer the question.

When both sides are in the same group, seek guidance for x, and seek clear reputation y+xy'-eˣ+eʸy'＝0y'dy dx (e -y) (x+e) When x answers or paragraph 0, y 0 is brought into the above equation dy dx 1

Question 3, Question 2. Thank you.

Summary. Thank you for the title.

Thank you for the title.

Answer the first question of the third question.

Hello dear after the query results are shown as the next Fuye: according to the Taylor Gong Leak Cluster This question is the limit of 1 2, I wish you a happy life and academic success. <>

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If you have this problem, it's actually a dead calculation, and then you have to pay attention to it, when you start to find out that x has two values, one of them should be rounded, because this is an implicit function, and it is impossible for x=0 to be meaningless, and then the calculation should not make mistakes, it should be b!

If you think what I'm saying makes sense, you can take a look.

Solution: Derivation on both sides.

3y²dy/dx=1+(1+dy/dx)*(1/sqrt(1-(xy)²)

Simplified after the item is moved.

3y²dy/dx+(1/sqrt(1-(xy)²)dy/dx=1-(1/sqrt(1-(xy)²)

y'=dy/dx=【1-(1/sqrt(1-(xy)²)3y²+(1/sqrt(1-(xy)²)

1.Differential with implicit functions.

Let f[x,y,z] = z -3xyz-a

z'x = -f'x/f'z = yz/(z²-xy)

z'y = -f'y/f'z = xz/(z²-xy)

z is also a function of y, and I just threw it as a constant - -

z''xy = [z'x]'y = [(yz)'(z² -xy) -yz * 2z z'y - x)]/(z²-xy)²

(z + y z'y)(z²-xy) -2yz² z'y + xyz]/(z²-xy)²

z³ -yz² z'y - xy² z'y)/(z²-xy)²

z³ -yz²+xy²)xz/(z²-xy)]/(z²-xy)²

z(z^4 - 2xyz³ -x²y²z)/(z²-xy)³

The result is too complicated, and I won't write specific about the steps for you.

The first-order partial derivative is about x, and we continue to regard the first-order partial derivative of the place as a function about z, and continue to find the partial derivative of y for this function. That's where the first question comes out. During this period, you can find the first-order partial derivative and then find the position z'

The second question is the same as the first question, finding the formula for the first-order partial derivative, just find the first-order partial derivative for x.

<>1. For the question of implicit functions for high numbers, the solution process is shown in the figure above.

2. When finding the derivative of the implicit function, first construct f, and then, use the implicit function to find the formula, that is, the formula of the second row, to find the derivative of the implicit function.

See above for the detailed steps to solve the problem of implicit functions for higher numbers.

It's about finding a derivative of x on both sides!

Note that y is a function of x.

E y is a composite function, first the derivative of e y as a whole, and if it is an exponential function, the derivative is e y, and then the derivative of y is y'

In the same way, xy is the derivative, =y+xy'

So ......

In the process of solving, when x 0 is first found, the function value y 1; is foundFind the value of the derivative function corresponding to y' when x 0 is found in turn.

To be continued. Repeat the above process to find the value of the second derivative at x 0.

For reference, please smile.

The equation e y+xy=e determines y=y(x); Seek dy dx ; d²y/dx²；

Solution 1: Derive x directly on both sides of the equation. Note in this way: e y is a function of y, and y is a function of x, so e y

When finding the derivative of x, we should use the chain rule of the composite function, i.e., d(e y) dx=[d(e y) dy][dy dx)=(e y)y'；

Again, where xy is a function of x and y, d(xy) = (dx dx)y+x(dy dx)=y+xy';

There is a formula where you draw a red line: (e y) y'+y+xy'=0；∴y'=-y/(x+e^y)

Derivative again: (e y) (y')²+e^y)y''+y'+y'+xy''=0, i.e. there is (e y)(y')²+e^y)y''+2y'+xy''=0.

y''=-[(e^y)(y')²+2y']/(x+e^y);

Put the y that has been found above'Substitution, that is:

y''=-[(e^y)y²/(x+e^y)²-2y/(x+e^y)]/(x+e^y)=[-(e^y)y²+2y(x+e^y)]/(x+e^y)³

2y(x+e^y)-(e^y)y²]/(x+e^y)³;

Solution 2: Derivation formula with implicit function:

Let f(x,y)=e y+xy-e=0, then:

y'=dy/dx=-(∂f/∂x)/(∂f/∂y)=-y/(x+e^y);

Note: The above is to find the partial derivative, x and y are on equal standing, don't use the chain rule again.

d²y/dx²=dy'/dx=[-(x+e^y)y'+y(1+y'e^y)]/(x+e^y)²

This is to find the full derivative, so continue to use the chain rule, the same as before].

2y(x+e^y)-(e^y)y²]/(x+e^y)³;

Replace the y that has been found above'Substitution and simplification. 】