How do you find the second derivative of #y^2 + x + sin y = 9#?

2 Answers
Aug 4, 2015

The second derivative is #y''=(sin(y)-2)/(2y+cos(y))^3#

Explanation:

Implicit differentiation gives us

#d/dx(y^2)+d/dx(x)+d/dx(sin(y))=d/dx(9)#
#2y dy/dx+1+cos(y) dy/dx=0#
#2y dy/dx+cos(y) dy/dx=-1#
#dy/dx(2y +cos(y))=-1#. Therefore,

#y'=dy/dx=(-1)/(2y +cos(y))#. This implies that #2y+cos(y)!=0#.

Taking the second implicit derivative gives us
#d/dx[dy/dx(2y +cos(y))=-1]#

By application of the Product Rule, we have
#d/dx(dy/dx)(2y +cos(y))+dy/dx(2dy/dx-sin(y) dy/dx)=0#
#(d^2y)/dx^2(2y +cos(y))+2((dy)/dx)^2-sin(y) ((dy)/dx)^2=0#
#(d^2y)/dx^2(2y +cos(y))=sin(y) ((dy)/dx)^2-2((dy)/dx)^2#
#(d^2y)/dx^2=(sin(y) ((dy)/dx)^2-2((dy)/dx)^2)/((2y +cos(y)))# or more succinctly

#y''=(y')^2(sin(y)-2)/(2y+cos(y))#
#y''=((-1)/(2y +cos(y)))^2(sin(y)-2)/(2y+cos(y))#
#y''=1/(2y +cos(y))^2(sin(y)-2)/(2y+cos(y))#
#y''=(sin(y)-2)/(2y+cos(y))^3#

Aug 4, 2015

I get: #y'' = (-2+siny)/(2y+cosy)^3#

Explanation:

#x+y^2+siny = 9#

#d/dx(x+y^2+siny) = d/dx(9)#

#1 + 2y dy/dx + cosy dy/dx = 0#

#dy/dx = (-1)/(2y+cosy) = -(2y+cosy)^-1#

#d/dx(dy/dx) = 1(2y+cosy)^-2*[d/dx(2y+cosy)]#

# = (2y+cosy)^-2 [2 dy/dx - siny dy/dx]#

# = (2y+cosy)^-2 [2 - siny] dy/dx#

# = (2y+cosy)^-2 [2 - siny] [-(2y+cosy)^-1]#

# = -(2-siny)(2y+cosy)^-3#

# = (-2+siny)/(2y+cosy)^3#