How do you use implicit differentiation to find #(d^2y)/(dx^2)# given #4y^2+2=3x^2#?

1 Answer
Mar 17, 2017

#(d^2y)/(dx^2)=3/(8y)[2-(3x^2)/(2y^2)]#

Explanation:

Implicit differentiation is a special case of the chain rule for derivatives.

Generally differentiation problems involve functions i.e. #y=f(x)# - written explicitly as functions of #x#. However, some functions y are written implicitly as functions of #x#.

So what we do is to treat #y# as #y=y(x)# and use chain rule. This means differentiating #y# w.r.t. #y#, but as we have to derive w.r.t. #x#, as per chain rule, we multiply it by #(dy)/(dx)#.

Hence implicit differentiating #4y^2+2=3x^2#, we get

#4xx2yxx(dy)/(dx)+0=3xx2x#

or #8y(dy)/(dx)=6x# ........................(1)

and #(dy)/(dx)=(6x)/(8y)=(3x)/(4y)# ........................(2)

Now implicitly differentiating (1) further, we get

#8(dy)/(dx)xx(dy)/(dx)+8yxx(d^2y)/(dx^2)=6#

or using (2) we get #8((3x)/(4y))^2+8y(d^2y)/(dx^2)=6#

or #8(9x^2)/(16y^2)+8y(d^2y)/(dx^2)=6#

or #(d^2y)/(dx^2)=1/(8y)[6-(9x^2)/(2y^2)]=3/(8y)[2-(3x^2)/(2y^2)]#