If we know that 2y-3 is proportional to 3x + 1, then the analytic expression of Y and X may be () A. y=3x+1B. y＝32x+1C. y＝32x+2D. y=3x+2

If we know that 2y-3 is proportional to 3x + 1, then the analytic expression of Y and X may be () A. y=3x+1B. y＝32x+1C. y＝32x+2D. y=3x+2

∵ 2y-3 is proportional to 3x + 1, then 2y-3 = K (3x + 1). When k = 1, 2y-3 = 3x + 1, that is, y = 32x + 2

Under the root sign of Z = arctan, x ^ y obtains the first order partial derivatives of X and Y respectively

dz/dx=y*x^(y/2-1)/2(1+x^y)
dz/dy=lnx*x^(y/2)/2(1+x^y)

Z = e ^ (UV) u = ln [radical (x ^ 2 + y ^ 2)] v = arctan (Y / x)
Partial derivative or derivative of compound function

U = ln [radical (x ^ 2 + y ^ 2)] = 1 / 2ln (x ^ 2 + y ^ 2) Z'x = ve ^ (UV) * 1 / [2 (x ^ 2 + y ^ 2)] * 2x + UE ^ (UV) * 1 / (1 + y ^ 2 / x ^ 2) * (- Y / x ^ 2) = ve ^ (UV) * x / (x ^ 2 + y ^ 2) - ue ^ (UV) * y / (x ^ 2 + y ^ 2) (U, V substitute by itself) Z'y = ve ^ (UV) * 1 / [2 (x ^ 2 + y ^ 2)] * 2Y + UE ^ (UV) * 1 / (1 + y ^ 2 / X