Implicit Differentiation

 

 

Motivation:

                     Previously,  the derivative function of  the form y=f(x) has been introduced. But in some occasions,  the relationship between y and x is not expressed in the form y=f(x).  Instead,  they might be expressed as  g(x,y)=c . But we still want to find out the derivative from this kind of situations. For example,

 

                                                x² + y² = 1  

 

                    We know this equation stands for a circle in the xy-plane. Given a point on the circle, the value of    at that point  is the slope of the tangent line passing that point.  In this case, we can express the curve as the combination of the two functions:

 

                                         y=     and   y=-

 

                    Thus,   can be found according to where the point is located.  But is there any way that we can directly use the original relationship to find out  without using the form y=f(x) ?  Yes, it is known as the differentiation of implicit functions. The basic idea is just using “chain rule”

 

 

 

 

Example ( Ellipse and Circle ):

                                           .  Find   .

     Sol:

                               Differentiate with respect to x on both sides,  we have

 

                                         

 

                             So, we know that   satisfies the relationship above. Once (x,y) is given,   at that point can be quickly determined .

 

 

 

 

Example ( Parabola ):

                                               x=y² + by +c, where b  0 .  Find   at (c,0) .

              Sol:

                                   Differentiate with respect to x on both sides:

 

                                         1=2y  +b

                                      1= (2y+b)

                                      So,        =

 

 

 

 

 

 

 

Example:   x³ + y³ = 4xy .  Find  at  (2,2) .

 

             Sol: 

                                     3x² + 3y²  =4x

 

                                      = -3

                         Thus, the tangent line at (2,2) is   y=-3(x-2)+2