Maths problems!!!

Q.2
b^2x^2 + a^2y^2 = a^2b^2
b^2(2x) + a^2(2y)(dy/dx) = 0
dy/dx = -2b^2(x)/2a^2(y) = -b^2(x)/a^2(y)
d2y/d2x =[ a^2y(-b^2) - (-b^2(x))(a^2(dy/dx))]/a^4y^2 = [-a^2b^2y + a^2b^2(x)(dy/dx)]/a^4y^2 = a^2b^2[-y + x(-b^2(x)/a^2(y)]/a^4y^2
= (b^2/a^2y^2)[-y - b^2x^2/(a^2y)]= -(b^2/a^4y^3)[a^2y^2 + b^2x^2]
=-(b^2/a^4y^3)(a^2b^2) = -b^4/(a^2y^3).

y= xe^(x^2)
dy/dx = e^(x^2) + x[e^(x^2)(2x)] = e^(x^2)[1 + 2x^2]
d2y/d2x = 4xe^(x^2) + ( 1 + 2x^2)[e^(x^2)(2x)] = e^(x^2)[4x + 2x(1+ 2x^2)]
= 2xe^(x^2)[2 + 1 + 2x^2] = 2x(3 + 2x^2)e^(x^2).
Q.3
2x + y = 500 and
Area, A= xy= x(500 - 2x) = 500x - 2x^2
dA/dx = 500 - 4x. Put it =0, therefore, 4x = 500, therefore, x = 125, so y = 250.
So for maximum area A, x = 125 and y = 250.

2008-06-26 13:09:36 補充：
Q.1 (sinx - sina)/(x-a) = [2cos(x+a)/2sin(x-a)/2]/(x-a) = cos(x+a)/2[(sin(x-a)/2)/(x-a)/2].
But lim x tends to a for [sin(x-a)/2/(x-a)/2] = 1. Therefore, lim x tends to a for the whole expression = cos(a +a)/2 = cosa.