There are lots of problems in Calculus in which we have to find out rate of change of variables which basically related to common variable. To solve that kind of problems we have to find appropriate and exact rate of change through Implicit Differentiation respect to time. The nature of rate of change is dependent on the dependent variable. Rate of change is positive if dependent variable raises with respect to time and rate of change is negative if dependent variable decreases with respect to time.
Now we are going to derive the rate of change formula calculus. In mathematics, all behavior of variables is shown through independent and dependent dimensions. The function of any graph describes a horizontal change of independent variable with respect to vertical change of dependent variable. So graph show changes according to time.
Now we have to find out where the function f(x) is changes between two points x1 and x2. For that we first have to put these values in independent variable 'x' after that we calculate the difference between the dependent variable,
So Δx = x2 – x1 …..........equation 1.
Where ‘Δx’ is a change in independent variable.
Same as above corresponding change in dependent variable
Δf = f(x2) – f(x1) ….............equation 2
Now from equation 1
x2 - x1 = Δx,
After moving ‘x1’ term right hand side
x2 = x1 + Δx,
Now we put the value of ‘x2’ in the equation 2
Δf = f(x1 + Δx) – f(x1),
Where Δx = x2 – x1,
Now to understand that more deeply we take an example of function f(x) where f(x) = 3. x3 and x1 , x2 are respectively 3, 5.
So Δx = 5 - 3 = 2,
Δf = f( x1 + Δx) - f(x1),
Δf = f(3 + 2) - f(3),
Δf = 3*53 - 3*33 =294,
Δf = 294.