Implicit Differentiation
Key Questions

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^2x^2 cdot 2y{dy}/{dx}}/{(y^2)^2} ={2x(y^2xy{dy}/{dx})}/{y^4}#
by plugging in#{dy}/{dx}={x^2}/{y^2}# ,
#Rightarrow{d^2y}/{dx^2}={2x[y^2xy(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# 
As the first step, we will differentiate once, and apply the product rule:
#d/dx[x^3]*y^3 + d/dx[y^3]*x^3 = d/dx[8]# For
#y^3# , remember to use the chain rule. Simplifying yields:#3x^2y^3 + 3y^2x^3dy/dx = 0# Now, we will solve for
#dy/dx# :#dy/dx = (3x^2y^3)/(3y^2x^3)# We can cancel off the 3, an
#x^2# , and a#y^2# , which will yield:#dy/dx = y/x# Now, differentiate once again. We will apply the quotient rule:
#(d^2y)/(dx^2) = (x*dy/dx  y*1)/x^2# Looking back at the previous equation for
#dy/dx# , we can substitute into our equation for the second derivative to get it in terms of only#x# and#y# :#(d^2y)/(dx^2) = (x*(y/x)  y*1)/x^2# Simplifying yields:
#(d^2y)/(dx^2) = (2y)/x^2# 
Implicit differentiation is a way of differentiating when you have a function in terms of both x and y. For example:
#x^2+y^2=16# This is the formula for a circle with a centre at (0,0) and a radius of 4
So using normal differentiation rules
#x^2# and 16 are differentiable if we are differentiating with respect to x#d/dx(x^2)+d/dx(y^2)=d/dx(16)# #2x+d/dx(y^2)=0# To find
#d/dx(y^2)# we use the chain rule:#d/dx=d/dy *dy/dx# #d/dy(y^2)=2y*dy/dx# #2x+2y*dy/dx=0# Rearrange for
#dy/dx# #dy/dx=(2x)/(2y# #dy/dx=x/y# So essentially to use implicit differentiation you treat y the same as an x and when you differentiate it you multiply be
#dy/dx#