Question

Check my work please:
dy/dx of Ln[x*y^2]=24 is -2x/y

Answer 1

dy/dx = -2xy

Answer 2

i got something really different

are you trying to find y' ?

Answer 3

d/dx (xy^2) = d/dx(y), (x)(2y(dy/dx)) + (y^2)(1) = 0, 2xy(dy/dx + y^2 = 0, 2xy(dy/dx) = -y^2

Answer 4

oops in the first part, d/dx(xy^2) = d/dx(24)* my bad

Answer 5

\[\ln (xy^{2}) = 24\]find dy/dx

Answer 6

thats it

Answer 7

ur answer is -2xy

Answer 8

what did you do with the natural log Ln ?

Answer 9

im sry, so stupid

Answer 10

forgot it

Answer 11

hehe it happens

Answer 12

thank you MathMind

Answer 13

-y/ 2x

Answer 14

thats the correct answer