Implicit Differentiation

Implicit Differentiation

An example of a function that is explicitly defined would be y=f(x).

But unfortunately not all functions are explicitly defined, such as the equation of a circle x^2 + y^2 = r^2. This is an example of a function that is implicitly defined. Often with implicitly defined functions it is not possible to solve for y.
             
            To equation of a circle can be divided into two pieces: the top half of the circle and the bottom half of the circle, with         
            each piece a separate function 

The Implicit Differentiation Process
      1. Differentiate both sides of the equation with respect to x
      2. Collect the terms with dy/dx on one side of the equation
      3. Factor out dy/dx
      4. Solve for dy/dx

example 1:  

               x^3 + y^3 - 9xy = 0   

The graph of x^3 + y^3 - 9xy = 0, a folium


                                                       
-This function has well defined tangent lines at every point on the graph. 
-Even though it is not a function in the classic sense, it is still differentiable
-The derivative from the right = the derivative from the left

x^3 + y^3 -9xy = 0 

3x^2 + 3y^2 * dy/dx - 9x*dy/dx -9y = 0           Since y is a function of x, when taking the derivative of any term with y,              
                                                                           multiply by dy/dx   
3y^2*dy/dx - 9x*dy/dx = 9y - 3x^2

y^2*dy/dx - 3x*dy/dx = 3y - x^2

dy/dx( y^2 - 3x ) = 3y -x^2

dy/dx = (3y - x^2)/(y^2 -3x)



*can use notation y' or dy/dx


example 2: 
                         2y = x^2 + siny

2*y' = 2x + cosy*y'

2*y' - cosy*y' = 2x

y'( 2 - cosy) = 2x

y' = 2x / (2 - cosy)


Finding the tangent and normal lines to a curve:
      -to find the slope of the tangent at a point, plug the coordinates of the point into the equation for dy/dx


example 1:
                         x^2 - xy + y^2 = 7  at ( -1,2 )



2x - x*dy/dx - y + 2y*dy/dx = 0

-x*dy/dx + 2y*dy/dx = y - 2x

dy/dx ( 2y-x ) = y -2x

dy/dx = ( y- 2x )/( 2y-x )

dy/dx(-1,2) = (2 - 2* -1)/(2*2 + 1) = 4/5

tangent: y-2 = 4/5(x+1)
normal: y-2 = -5/4(x+1)        
*To find slope of normal line, use opposite inverse of slope of tangent line


In class, we took our first homework quiz over the derivatives thus far, with the exception of the exponential and logarithmic derivatives

Homework: page 155# 3-28, multiples of 3