IMPLICIT DIFFERENTIATION PROBLEMS


The following problems require the use of implicit differentiation. Implicit differentiation is nothing more than a special case of the well-known chain rule for derivatives. The majority of differentiation problems in first-year calculus involve functions y written EXPLICITLY as functions of x . For example, if

$ y = 3x^2 -\sin(7x+5) $ ,

then the derivative of y is

$ y' = 6x - 7 \cos(7x+5) $ .

However, some functions y are written IMPLICITLY as functions of x . A familiar example of this is the equation

x2 + y2 = 25 ,

which represents a circle of radius five centered at the origin. Suppose that we wish to find the slope of the line tangent to the graph of this equation at the point (3, -4) .

How could we find the derivative of y in this instance ? One way is to first write y explicitly as a function of x . Thus,

x2 + y2 = 25 ,

y2 = 25 - x2 ,

and

$ y = \pm \sqrt{ 25 - x^2 } $ ,

where the positive square root represents the top semi-circle and the negative square root represents the bottom semi-circle. Since the point (3, -4) lies on the bottom semi-circle given by

$ y = - \sqrt{ 25 - x^2 } $ ,

the derivative of y is

$ y' = - (1/2) \big( 25 - x^2 \big)^{-1/2} (-2x) = \displaystyle{ x \over \sqrt{ 25 - x^2 } } $ ,

i.e.,

$ y' = \displaystyle{ x \over \sqrt{ 25 - x^2 } } $ .

Thus, the slope of the line tangent to the graph at the point (3, -4) is

$ m = y' = \displaystyle{ (3) \over \sqrt{ 25 - (3)^2 } } = \displaystyle{ 3 \over 4 } $ .

Unfortunately, not every equation involving x and y can be solved explicitly for y . For the sake of illustration we will find the derivative of y WITHOUT writing y explicitly as a function of x . Recall that the derivative (D) of a function of x squared, (f(x))2 , can be found using the chain rule :

$ D \{ ( f(x) )^2 \} = 2 f(x) \ D \{ f(x) \} = 2 f(x) f'(x) $ .

Since y symbolically represents a function of x, the derivative of y2 can be found in the same fashion :

$ D \{ y^2 \} = 2 y \ D \{ y \} = 2 y y' $ .

Now begin with

x2 + y2 = 25 .

Differentiate both sides of the equation, getting

D ( x2 + y2 ) = D ( 25 ) ,

D ( x2 ) + D ( y2 ) = D ( 25 ) ,

and

2x + 2 y y' = 0 ,

so that

2 y y' = - 2x ,

and

$ y' = \displaystyle{ - 2x \over 2y } = \displaystyle{ - x \over y } $ ,

i.e.,

$ y' = \displaystyle{ - x \over y } $ .

Thus, the slope of the line tangent to the graph at the point (3, -4) is

$ m = y' = \displaystyle{ - (3) \over (-4) } = \displaystyle{ 3 \over 4 } $ .

This second method illustrates the process of implicit differentiation. It is important to note that the derivative expression for explicit differentiation involves x only, while the derivative expression for implicit differentiation may involve BOTH x AND y .

The following problems range in difficulty from average to challenging.






Click HERE to return to the original list of various types of calculus problems.


Your comments and suggestions are welcome. Please e-mail any correspondence to Duane Kouba by clicking on the following address :

kouba@math.ucdavis.edu



Duane Kouba
1998-06-23