Solve the equation explicitly for y and differentiate to get y' in terms of x.
2006-10-05 10:05:13 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
9y^2 = 36-4x^2 by rearranging
y = 1/3(sqrt(36-4x^2))
y'= 1(/3)(1/2)(36-4x^2)^(-1/2) d/dx(36-4x^2)
d/dx(u^n) = nu^(n-1)du/dx
= 1/6((36-4x^2)^(-1/2)(-8x) = -4/3x(36-4x^2)(^1/2)
2006-10-05 14:23:23 · answer #1 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
d/dx[4x^2 + 9y^2] = d/dx[36]
8x + 18y*dy/dx = 0
dy/dx = -8x/18y
dy/dx = -4x/9y
2006-10-05 17:11:30 · answer #2 · answered by nckobra40 3 · 0⤊ 0⤋
y=[2x (plus or minus) 6] (divided by) [(plus or minus)3]
2006-10-05 17:11:19 · answer #3 · answered by Chrissy 2 · 0⤊ 0⤋
9y^2=36-4x^2
y^2=(36-4x^2)/9
y=rt[(36-4x^2)/9]
9y^2=36-4x^2
18yy'=-8x
y'=(-8/18)(x/y)
=(-4/9)(x/y)
2006-10-05 17:09:21 · answer #4 · answered by raj 7 · 0⤊ 0⤋