1
answer
0
watching
104
views
6 Nov 2019
Find the second derivative y'(x) of -3y^2=-2x^3+x-cos y
Find the second derivative, y"(x), of - 3y2 = - 2x3 + x - cos y. y"(x) = - 4+(- cos y - 3)(y')2/- 3y + sin y y"(x)= - 2x + (cos y - 6)(y')/- 6y - cos y y"(x) = - 12x + (cos y - 3)y2/- 6y2 - sin y y"(x) = - 12x + (cos y + 6)(y')2/- 6y - sin y Show transcribed image text
Find the second derivative y'(x) of -3y^2=-2x^3+x-cos y
Find the second derivative, y"(x), of - 3y2 = - 2x3 + x - cos y. y"(x) = - 4+(- cos y - 3)(y')2/- 3y + sin y y"(x)= - 2x + (cos y - 6)(y')/- 6y - cos y y"(x) = - 12x + (cos y - 3)y2/- 6y2 - sin y y"(x) = - 12x + (cos y + 6)(y')2/- 6y - sin y
Show transcribed image text Jean KeelingLv2
18 Aug 2019