Derivative Of Implicit Function

In this article let us learn what is derivative at a point and what do we  mean by implicit functions and how to find the derivative of implicit functions.
Derivative at a point:
Let f be any function defined in a domain and a be any point on the domain. Let "h" be any small arbitrary small positive or negative number. Definitely there will be a difference between the values f(a+h) and f(a). If the difference is divide by the small value "h" then the resulting number is called as difference quotient of f at a with the small increment h. Difference quotient is denoted as f(a+h). Actually this represents average rate of change of f on the given interval. As h tens to zero, a limiting value can be calculated ,
 i.e., `lim_(h->0)` `(f(a+h) - f(a))/(h)` which represents the rate of change of "f" at "a".
The derivative of a function f at a point x=a is defined by f'(a) = `lim_(h->0)` `(f(a+h) - f(a))/(h)`  (when limit exists)
There are many functions like exponential functions, logarithmic function, trigonometric functions and always there is a need to find the derivative of them. We could not always use the above formula to calculate the derivative. So for our convenience we have derived some standard results. Some of them are

Algebra is widely used in day to day activities watch out for my forthcoming posts on solving algebraic proportions and how to solve proportion problems. I am sure they will be helpful.

Learning about Derivative of Implicit Function
If any of the functions are given as an algebraic equation then such type of equations are called as implicit functions.
Example: xy = 8(x+y) As  we could not solve x in terms of y or y in terms of x, the algebraic equation is an implicit function.
Problems on Derivative of Implicit Function
Find `dy/dx` for the following implicit functions.
`sqrt(x)` + `sqrt(y)` = 100
   Differentiating w.r.t x, we have
          `(1)/(2sqrt(x))` + `(1)/(2sqrt(y))` = 0     refer the above table for derivative of `sqrt(x)`  and constant
sin x + log y = x2 + 18x + 3
   Differentiating w.r.t x, we have
      cos x + `(1)/(y)` `dy/dx` = 2x + 18
                     `dy/dx` =  y(2x + 18 - cos x)