Find dy/dx at the point (3,8) for 5xy^2/3-x^2y=-12
A.2
B.17
C.-7
D.-28/3
3 answers
What was your dy/dx ?
5 x y^(2/3) - x^2 y= -12 ???
5[x y^-(1/3) (2/3) dy + y^(2/3)dx] = x^2 dy + 2 x y dx
now plug in x = 3 and y = 8
5[ 1 dy + 4 dx ] = 9 dy -20 dx
5 dy - 9 dy = 48dx - 20 dx
-4 dy = -28 dx
dy/dx = -28/4 = -7
5[x y^-(1/3) (2/3) dy + y^(2/3)dx] = x^2 dy + 2 x y dx
now plug in x = 3 and y = 8
5[ 1 dy + 4 dx ] = 9 dy -20 dx
5 dy - 9 dy = 48dx - 20 dx
-4 dy = -28 dx
dy/dx = -28/4 = -7
I had to change the problem statement typing, figured it meant y^(2/3) not y^2 / 2
0 answers