微積分求解過程！
3 個解答
- 麻辣Lv 76 年前最佳解答
1.x^3*y^2 + 3x^4*y^4 = 3x^5(3x^2*y^2 + 2x^3*yy') + (12x^3*y^4 + 12x^4*y^3*y') = 15x^4y'= [15x^4 - 3(xy)^2 - 12x^3*y^4]/2(yx^3 + 6x^4*y^3)= (15x^2 - 3y^2 - 12xy^4)x^2/2(1 + 6xy^2)yx^3= (15x^2 - 3y^2 - 12xy^4)/2xy(1 + 6xy^2) 2.x^2 = (x + y)/(x - y) x + y = (x - y)x^2 = x^3 - yx^21 + y'= 3x^2 - x^2*y' - 2xyy'= (3x^2 - 2xy - 1)/(1 + x^2) 3.y = x^3*sin4x + x*cos4xy'= 3x^2*sin4x + 4x^3*cos4x + cos4x - 4x*sin4x= x(3x - 4)sin4x + (4x^3 + 1)cos4x 4.y = -2*asin(3x)y'= -2*3/√[1 - (3x)^2] = -6/√(1 - 9x^2)
Note: [asin(ax)]'= a/√[1-(ax)^2] 5.y = ln[√(3x^2 + 13x + 4)]y'= (3x^2 + 13x + 4)'/(3x^2 + 13x + 4)= (6x + 13)/(3x^2 + 13x + 4) 6.y = 2e^2x + e^(-3x)y'= 4e^2x - 3e^(-3x) 7.Limit(x->3)[sin(x - 3)/(x^2 - x - 6)]= [sin(x - 3)]'/(x^2 - x - 6)'.....羅必達法則= cos(x - 3)/(2x - 1)= cos0/(6-1)= 1/5
