![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
X = u + V y = u ^ 2 + V ^ 2 z = u ^ 3 + V ^ 3 finding the partial differential of Z to X
According to u ^ 3 + V ^ 3 = (U + V) (u ^ 2 + V ^ 2-uv), we can get z = x [y - (x ^ 2-y) / 2], and then we can get the partial derivative of X to get - 3UV
Total differential: z = x ^ 3-4x ^ 2 + 2xy-y ^ 2
z=x^3-4X^2+2XY-Y^2 ,
dz=(3x^2-8x+2y)dx+(2x-2y)dy.
RELATED INFORMATIONS
- 1. The meaning of differential, Derivative is the rate of change, integral is the magnitude of change, what about differential? In addition, what is the significance of studying differential?
- 2. Several concepts of advanced numbers, If the function f (x, y) has a minimum at point (x0, Y0), then (x0, Y0) must be a function of F (x, y) The minimum value of a continuous point B in the domain of definition C the minimum value of stationary point D in a certain field (x0, Y0) Let f (x), G (x) be continuous on [a, b], and f (x) ≥ g (x), then A ∫ (upper limit B, lower limit a) f (x) DX ≥ ∫ (B, a) g (x) DX B ∫(b,a)f(x)dx≤∫ (b,a)g(x)dx C ∫ f(x)dx≥∫ g(x)dx D ∫ f(x)dx=∫ g(x)dx What is the divergence of the following generalized integrals A ∫ (upper limit + ∞, lower limit 0) DX / 1 + x ^ 2 B ∫ (1,0) DX / radical 1-x ^ 2 C ∫(+∞,e)(lnx/x)dx D ∫(+∞,e)e^-x dx The continuity of function f (x, y) in P 0 (x 0, y 0) is the existence of the first partial derivative of F (x, y) in P 0 (x 0, y 0) A necessary condition B sufficient condition C necessary and sufficient condition D neither necessary nor sufficient condition Let f (x, y) = {(x ^ 2) y / x ^ 4 + y ^ 2 (x, y) not = (0,0); 0 (x, y) = (0,0) Alim (x, y) → (0,0) f (x, y) exists and f (x, y) is discontinuous at (0,0) B LIM (x, y) → (0,0) f (x, y) does not exist, f (x, y) is discontinuous at (0,0) C LIM (x, y) → (0,0) f (x, y) exists and f (x, y) is continuous at (0,0) D LIM (x, y) → (0,0) f (x, y) does not exist, f (x, y) is continuous at (0,0) Let a = ∫ (2,1) lnxdx, B = ∫ (2,1) | LNX | DX, then A a=b B a＞b C a＜b D a≥h
- 3. The concept of function is based on mapping, but mapping does not allow one to many. Why does the so-called multi valued function allow one to many? Based on your answers, I come to the conclusion that: Generally, function is the abbreviation of single valued function, which is built on the basis of mapping. Multi valued function is a different concept system from function, that is, function is not the general name of single valued function and multi valued function. Multi valued function is an independent concept, which is not built on the basis of mapping. Its definition stipulates that it can be one to many I don't know if I can say that
- 4. Find the derivative and differential of function y = x ^ 3tan ^ 2 (x ^ 2-4)
- 5. Given function FX satisfies f (x + y) = f (x) + F (y) for any XY ∈ R Find the value of 1 f (0) 2 F (x) is an odd function
- 6. Finding the total differential of Z = XY + (x / y) at point (1,2)
- 7. Finding the total differential of Z = in (XY) + e ^ (x + y ^ 2)
- 8. A calculus problem f (x + y, X-Y) = XY, then what are the partial derivatives of F for X and F for y?
- 9. Evaluation: x + 3Y divided into x-2y, 3x-5y = 0
- 10. 3x+2y+z=14 { x+y+z=10 2x+3y-z=1
- 11. Finding the general solution of differential function y "- 2Y = e ^ x That's the process of finding the derivative of Y
- 12. General solution of differential equation y "+ y = ex General solution of differential equation y "+ y = e ^ x
- 13. Differential equation.. Y "+ y = cosx for general solution
- 14. The general solution of Y "+ y = x + cosx
- 15. What is the general solution of the differential equation y '' + y '^ 2 = 2E ^ (- y)?
- 16. Finding the general solution of the differential equation XY '' + y '= 0
- 17. How to find the general solution of first order linear differential equation
- 18. Y '' + y = x ^ 2 differential equation Y '' + y = x ^ 2 differential equation
- 19. Finding the square plus one of SiNx given TaNx = 2
- 20. Given f (1 + SiNx) = 2 + SiNx + cos2x, find f (x) Let u = 1 + SiNx, then SiNx = U-1 (0 ≤ u ≤ 2), then f (U) = - U2 + 3U + 1 (0 ≤ u ≤ 2) So f (x) = - x2 + 3x + 1 (0 ≤ u ≤ 2) How does [f (U) = - U2 + 3U + 1] come from? Ask for advice!