Calculate Y' and Y'' at the point (2,6)?

Question

On the curve #x^2+y^2=4x+9y-22#

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Answer 1 · 2018-06-18T21:20:30

#y'=0#, #y''=-2/3#

Differentiating the given expression implicitly wrt #x#
#2x+2ydy/dx=4+9dy/dx+0#, i.e, #dy/dx[2y-9]=[4-2x]#

#y'=dy/dx=[4-2x]/[2y-9]# and substituting for the given values of #y# and #x#
#dy/dx=[4-2[2]]/[2[6]-9]# = #0/3 = 0# i.e, [#y'=0]#

For the second derivative,#y''#=#[d^2y]/dx^2# = #d/dx[[4-2x]/[2y-9]]#

Using the quotient rule to differentiate this expression ie, #d/dx[u/v]=[vdu/dx-udv/dx]/[v^2# where #u# and #v# are both functions of #x# we obtain,

#[[[2y-9][-2] - [4-2x][2]]dydx]/[2y-9]^2#, but we know that #dy/dx=0# at #[2, 6]#

therefore, #y''= -2[2y-9]/[2y-9]^2# = #-2/3# when #y=6#.