Seeking total differential

Seeking total differential

u=x^(yz)∂u/∂x = (yz)x^(yz-1)∂u/∂y = (zlnx)x^(yz)∂u/∂z = (ylnx)x^(yz)du = (∂u/∂x)dx + (∂u/∂y)dy + (∂u/∂z)dz=(yz)x^(yz-1)dx + (zln...

Let u = f (x, y) = ∫ (0 to XY) e ^ (- T ^ 2) DT find Du
The answer is Du = e ^ (- x ^ 2 * y ^ 2) (YDX + XDY)

du=∂u/∂xdx+∂u/∂ydy
=E ^ (- x ^ 2 * y ^ 2) &; (XY) / &; xdx + e ^ (- x ^ 2 * y ^ 2) &; (XY) / &; YDY (using the derivation of integral upper limit function)
=e^(-x^2*y^2)ydx+e^(-x^2*y^2)xdy
=e^(-x^2*y^2)(ydx+xdy)
I wish you progress in your studies~

Let u = (e ^ XY) (COS (x + y ^ 2)), find Du

We can't find the a of partial derivative Du / DX = Ye ^ XY (COS (x + y ^ 2)) + e ^ XY (- sin (x + y ^ 2)). We can't find the a of partial derivative Du / dy = Xe ^ XY (COS (x + y ^ 2)) + e ^ XY (- sin (x + y ^ 2)) 2ydu = {Ye ^ XY (COS (x + y ^ 2)) + e ^ XY (- sin (x + y ^ 2))} DX + {Xe ^ XY (COS (x + y ^ 2)) + e ^ XY (- sin (x + y ^ 2)) 2Y}

Let u = cosh (XY) + cos (XY), then Du=

What is the basis of using discriminant method to find the range of fractional function in mathematics of grade one in senior high school? Not how to find it, but why can I find it that way?

The domain of function is definitely not an empty set, so x must exist
After making a simple variation with the discriminant method, X still exists, that is, the quadratic equation of one variable must have a solution, that is, he of the discriminant is greater than or equal to 0. Thus, the value range of Y is obtained
It's not necessary to study why and know how to do it. Just like the substitution method, you can do it and get the desired result

On the problem of finding function range by discriminant method
For example, y = 50x / (1 + (the square of x)) with additional restrictions (x > 0) to find the maximum value of Y. among the answers given are discriminant method, simultaneous discriminant > = 0 and 50 / Y > 0. What does this 50 / Y > 0 mean? And how to use discriminant method to ensure that there is no error when the domain of definition is not r

Use the discriminant method to change the quadratic equation of X into YX ^ 2-50x + y = 0. Because the product of two is 1, it means that two have the same sign, so the sum of two is positive, that is, 50 / Y > 0. As for your second problem, there are two cases in which the domain of definition is not R. the first case is that one or two points have been cut out

How to find the value range of fractional function? It needs to be more general

1) If the numerator and denominator are all one time, y = (AX + b) / (Cx + D), then the direct division is: y = (AX + AD / C + B-AD / C) / (Cx + D) = A / C + (B-AD / C) / (Cx + D), so the range is Ya / C, (in the case of ADBC)
2) If the highest denominator is quadratic, y = (AX ^ 2 + BX + C) / (DX ^ 2 + ex + F), then the de denominator is transformed into a quadratic equation about X
x^2(y-a)+x(ey-b)+(fy-c)=0
According to delta = (ey-b) ^ 2-4 (Y-A) (fy-c) > = 0, the range of Y is obtained by solving this inequality
3) If the numerator denominator is higher than quadratic, other methods such as derivation are used

A method for calculating the range of fractional function

1. Separation constant method
y=(x^2+2)/(x^2-1)
=1+3/(x^2-1)
y1
2. Discriminant method
For the fraction whose numerator and denominator are quadratic functions
3. Slope formula method
The first degree formula of sine and cosine for numerator and denominator respectively
y=(sinx-a)/(cosx-b)
Consider the slope of the line between the moving point (cosx, SiNx) and the fixed point (B, a) on the unit circle
4. Derivative method
The range of fractional functions on [a, b] for polynomials whose numerator and denominator are x

Finding the range of function with fraction
Please explain by examples, hope to be easy to understand

1. Split term method: a fraction is reduced to the sum of several formulas, in which only one formula contains X. it is suitable for simple fraction function or fraction function whose numerator denominator x is once
Example: find the range of y = 2x / (5x + 1)
y=2[x+(1/5)-(1/5)]/5[x+(1/5)]=(2/5)-[2/5(5x+1)]
∵x≠-1/5 ∴y≠2/5
The value range is {y ︱ y ∈ R and Y ≠ 2 / 5}
2. Discriminant method: the original function is transformed into a quadratic equation of one variable with y as the coefficient about X. if the equation has a real root, △ ≥ 0, the value range of y can be obtained. It is suitable for fractional function with quadratic numerator denominator. (Note: in transforming the original function into a quadratic equation of one variable with y as the coefficient about X, X & sup2 should be considered; whether the coefficient is 0 or not)
Example: find the range of y = (2x & sup2; - 2x + 3) / (X & sup2; - x + 1)
The original function is reduced to (Y-2) x & sup2; - (Y-2) x + (Y-3) = 0
When y ≠ 0, the above quadratic equation with respect to X has real roots,
The solution is 2 ＜ y ≤ 10 / 3
When y = 2, the equation has no solution
The range of function is (2,10 / 3]
The above two methods are suitable for finding the range of fractional function

Find the range of a function fraction function
The formula is f (x) = x2 / (X-2) because the X of the molecule is square, which is different from the denominator
Please teach me the solution of the relevant formula,

f(x)=(x²-4+4)/(x-2)
=(x²-4)/(x-2)+4/(x-2)
=x+2+4/(x-2)
=(x-2)+4/(x-2)+4
x>2,x-2>0
Then (X-2) + 4 / (X-2) > = 2 √ [(X-2) * 4 / (X-2)] = 4
So f (x) > = 4 + 4 = 8
x0
Then f (x) = - [(2-x) + 4 / (2-x)] + 4
(2-x)+4/(2-x)>=2√[(2-x)*4/(2-x)]=4
-[(2-x)+4/(2-x)]