## Instantaneous Rate of Change Formula :

Instantaneous function of f(a) is equal to the lim Δx* *→0 and delta y axis and divided to section for s_{(}_{J mol-1)} in joule mol^{-1} is equal to the lim x axis lim Δx* *→0 and time taken0 t_{(s)} in second and multiply the Angular acceleration α_{(rad/s}^{2})in radians per second squared addition of height of subtraction time taken t_{(s)} in second and area of a_{(m}^{2}_{)}) in meter^{2} and division of height of immersed part h_{(cm) } in center meter .Hence the instantaneous rate of change formula can be written as

**F(a)= lim Δx →0 Δy/Δs _{(}_{J mol-1)} = lim Δx →0 t_{(s)}(α_{(rad/s2)}+h_{(cm)})-(t_{(s)}(a_{(m2)})/h_{(cm)}**

Where,

Δs→ section in J mol^{-1}

T→ time taken in second

α→ Angular acceleration in radians per second squared

h→ height of the immersed part in cm

A → area in m2

With respect to x when x=a and y=f(x)

## Sample Problems

### Problem 1:

Calculate the Instantaneous rate of change of the function f(x) = 4x^{2} + 18 at x = 6 ?

### Solution :

Known Function

Y= f(x) = 4x^{2} + 18

F(x)=4(2x)+0

F(x)=8

Thus the instantaneous rate of change x=26

F(6)=8(6)

F=48

### Problems 2:

Calculate the Instantaneous rate of change of the function f(x) = 6x^{3} – 2x^{2} + 2x + 1 at x = 3?

### Solution :

Known Function

Y= f(x) = 6x^{3} – 2x^{2} + 2x + 1

f(x) = 6(3x^{2}) – 2(2x) + 2+ 0

f(x) =18x^{2}-4x+2

Thus the instantaneous rate of change at x=3

f(x) =18(3)^{2}-4(3)+2

=162-12+2=152

F(x)=152