Word Problems (Cubic Function) Mathematics N3

 

1.13 Implicit Differentiation                                                                                                                                    

The table below illustrates the concept of implicit differentiation. The equations of the straight line, quadratic functions, cubic functions and the exponential functions are very example of explicit differention.

For these functions the expression on the right hand    side are all written in terms of x. The subject of the formula is y. In order to calculate any specific value of y, the corresponding x value of that function is substituted into the function.

This is not the case for the equations of the rectangular hyperbolic functions, circle, ellipse and the hyperbola.

For the rectangular hyperbola, k=xy, the conversion of this function into the form y=k/x is necessary.                                               

The differentiation of the functions in which y is the subject of the formula is known as explicit differentiation. Implicit differentiation is applied to differentiate problems in which:                                                                                                            

a) it is not necessary to make y the subject of the formula.                                                                                           

b) it is very difficult to make y the subject of the formula

Table 1.23                                                                                                                                                                         

Table 1.23 : Implicit Differentiation
		Formula	Explicit	Implicit
1.	Straight Line	y=2x+7	y=2x+7
2.	Quadratic Functions	y= -(3x+5)2	y= -(3x+5)2
3.	Cubic Function	y= -3x3	y= -3x3
4.	Rectangular Hyperbola	K=xy	y=k/x	10=xy
5.	Hyperbola	1
6.	Circle	x2+y2=r2		x2+y2=81
7.	Ellipse	1

 

Example 1   Differentiate implicitly the following:                                                                  

a) 10=xy                                                                                                            

b)  x²+y²=81                                                                                                  

c) 1                                                                                               

d) 1

a)      The differentiation of the term xy is based on the product rule. In order to differentiate y, differentiate with respect to y and multiply it with  

  Mathematically, = + (y            (RHS)

Therefore,             = [(x). X  

                                = [(x). .           (RHS)

                                = x  

The derivative of 10 or a constant term is zero. The differential of the LHS is zero.                                                  Therefore,

= x  

x =

= -

b)      The derivative of 81 or a constant term is zero.                                                   

[2x] + [ X    = 0

          2x+ [2y    = 0

Therefore,       2y.  = -2x

   = 

       = 

 

 

c)   The derivative of 1 or a constant term is zero.                                                  

[  2x] - [   X    = 0

             [ ] - [  2y    = 0

Therefore,             =  X

   = 

d)   The derivative of 1 or a constant term is zero.                                                   

[  2x] + [   X    = 0

             [ ] + [  2y    = 0

Therefore,             =  X

   = -