How do you find the second derivative by implicit differentiation?

1 Answer
Wataru
Sep 12, 2014

Let us find #{d^2y}/{dx^2}# for #x^3+y^3=1#.

First, let us find #{dy}/{dx}#.
#x^3+y^3=1#
by differentiating with respect to #x#,
#Rightarrow 3x^2+3y^2{dy}/{dx}=0#
by subtracting #3x^2#,
#Rightarrow3y^2{dy}/{dx}=-3x^2#
by dividing by #3y^2#,
#Rightarrow {dy}/{dx}=-{x^2}/{y^2}#

Now, let us find #{d^2y}/{dx^2}#.
by differentiating with respect to #x#,
#Rightarrow{d^2y}/{dx^2}=-{2x cdot y^2-x^2 cdot 2y{dy}/{dx}}/{(y^2)^2} =-{2x(y^2-xy{dy}/{dx})}/{y^4}#
by plugging in #{dy}/{dx}=-{x^2}/{y^2}#,
#Rightarrow{d^2y}/{dx^2}=-{2x[y^2-xy(-x^2/y^2)]}/y^4=-{2x(y^2+x^3/y)}/y^4#
by multiplying the numerator and the denominator by #y#,
#Rightarrow{d^2y}/{dx^2}=-{2x(y^3+x^3)}/y^5#
by plugging in #y^3+x^3=1#,
#Rightarrow{d^2y}/{dx^2}=-{2x}/y^5#

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