Differentiation of Implicit Functions

In calculus, a method called implicit differentiation makes use of the chain rule to differentiate implicitly defined functions. As

explained in the introduction, y can be given as a function of x implicitly rather than explicitly. When we have an equation R(x, y) = 0.However,

sometimes it is simpler to differentiate R(x, y) with respect to x and then solve for `dy/dx` .

                                                                                                                                                                                                           

 

Differentiation of implicit function:

 

          The differentiation of implicit function theorem can be defined as if F is an open disk containing (p, q), where F(p, q) = 0, Fy(p,q)`!=` 0, and Fxy are continuous on the disk, then the equation F(x, y) = 0 defines y as a function of x near the point (p,q)and the derivative of this function is, and F

                       Differentiation of y with respect to x = ` -((((delF)/(delx)))/(((delF)/(dely))))` =` -(F_x/F_y)`

                        Differentiation of y with respect to x : `dy/dx` 

                        The differentiation of F with respect to var can be expressed as Fvar

          Total differentiation of implicit function with respect to x of both sides of F(x, y) = 0.

                                ` (delF)/(delx)` ` (dx)/(dx)` + ` (delF)/(dely)`` (dy)/(dx) ` = 0 = `(delF)/(delx)` + `(delF)/(dely)` `(dy)/(dx)`

 

 

Differentiation of implicit function problems:

 

Differentiation of implicit function problem 1:

              Find the differentiation of implicit function.sin (2x + y) =  8x

    Solution:

                                 Given implicit function,  sin ( 2x + y ) =  8x

                    Differentiation of Implicit function with respect to x,

                                                     `d/dx` sin(2x + y)  =  `d/dx` 8x

                                       cos ( 2x + y ) . {2 + `dy/dx` } = 8

                                                                 2 + `dy/dx` = `8 / cos ( 2x + y )`

                                                                       ` dy/dx` = `8 / cos ( 2x + y ) - 2`

                                                                       ` dy/dx` = ` 8 sec ( 2x + y ) - 2`

                                                                        ` dy/dx ` = ` (8 sec ( 2x + y ) - 2)`

    Answer:  The differentiation of implicit function sin(2x + y) =  8x is,   ` dy/dx ` = 8 sec(2x + y ) - 2

 

Differentiation of implicit function problem 2:

          Find the differentiation of implicit function..  x = cos (yx)

   Solution:

                            Given implicit function,   x = cos (yx)

                Differentiation of Implicit function with respect to x,

                                            ` d/dx` x = `d/dx ` (cos yx)

                                                    1 = - sin yx ( y + x`(dy)/(dx)` )

                                `-1/sin (yx)` = ( y + x`(dy)/(dx)` )

                                    - cosec yx = ( y + x`(dy)/(dx)` )

                               - cosec yx - y =  x`(dy)/(dx)`

                ` -(1/x)` (y + cosec yy) = `(dy)/(dx)`

     Answer: The differentiation of implicit function x = cos xy is,  `(dy)/(dx)` =  ` -(1/x)` (y + cosec yx)

Differentiation of implicit function problem 3:

            Find the differentiation of implicit function..   xy = sin y

     Solution:

                            Given implicit function,  xy = sin y

                Differentiation of Implicit function with respect to x,

                                                           ` d/dx` (x y)  = `d/dx ` (sin y)

                                                    y + x ` (dy/dx)`   = cos y (`(dy)/(dx)` )

                               y + x(`(dy)/(dx)`) - cos y `(dy/dx)` = 0

                                       y + (x - cos y) `(dy/dx) ` = 0

                                              (x - cos y)`(dy/dx)` = - y

                                                                `(dy/dx)` = `(- y)/(x - cos y)`

     Answer: The differentiation of implicit function xy = sin y is,    `(dy/dx)` = `(- y)/(x - cos y)`