The rate of change of quantities can be expressed in the form of derivatives. Rate of change of one quantity with respect to another is one of the major applications of derivatives. The rate of change of a function with respect to another quantity can also be done using chain rule.
Rate of change is usually defined by change of quantity with respect to time. For example, the derivative of speed represents the velocity, such that ds/dt, shows rate of change of speed with respect to time. Another example is the rate of change of distance with respect to time.
Similarly, if a quantity, say y changes with respect to another quantity, x, such that, y = f(x), then the rate of change of y with respect to x, is given by dy/dx of f’(x).
Hence, for the function, y = f(x), d/dx f(x), represents the rate of change of y with respect to x. Or we can say, dy/dx is the derivative of y with respect to x.
Note: If the rate of change of quantity, dy/dx, increases, then it is represented by positive sign If the rate of change of quantity, dy/dx, decreases, then it is represented by negative sign 

Chain Rule For Rate of Change of Quantities
If two variables, say x and y, changes with respect to another variable t, i.e., if x = f(t) and y = f(t) then according to the Chain Rule:
dy/dx = (dy/dt)/(dx/dt)
Where, dx/dt≠0
Hence, we can calculate the rate of change of variable y with respect to x, by finding the rate of change of both y and x with respect to t.
Derivative Related Articles 

Solved Examples
Q.1: If the radius of a circle is r = 5cm, then find the rate of change of the area of a circle per second with respect to its radius.
Solution: Given,
Radius of a circle =5cm
We know that,
Area of a circle, A = πr^{2}
Therefore, the rate of change of the area A with respect to its radius r will be:
dA/dr = d(πr^{2})/dr
= 2πr
By putting the value of given radius, r=5cm, we get;
dA/dr = 2π(5)
= 10π
Hence, the area of the circle is changing at the rate of 10π cm^{2} per second.
Example 2: Volume of the cube is increasing at a rate of 9 cubic inches per second. What is the rate at which surface area is increasing when the length of the edge of the cube is 10 inches?
Solution: Given,
Increasing rate of volume of cube = 9 cubic inches per second
Let the length of the edge of cube = a inches
By the volume of cube, we know that,
Volume of a cube = a^{3}
V = a^{3}
Differentiating both the sides, we get the rate of volume,
dV/dt = da^{3}/dt
9 = (da^{3}/da) . (da/dt)
9 = 3a^{2} .da/dt
9/3 = a^{2}. da/dt
da/dt = 3/a^{2} ……….(1)
Now, Surface area of cube, S = 6a^{2}
Rate of change of surface area will be:
dS/dt = d(6a^{2})/dt
dS/dt = d(6a^{2})/da . da/dt
dS/dt = 12a.(3/a^{2}) = 36/a
Since, new edge length = 10 inches
dS/dt = 36/10 = 3.6 sq.inches/sec
Practice Questions


Frequently Asked Questions on Rate of Change of Quantities
What is the rate of change of quantity?
When a quantity changes with respect to time, then it is said to be the rate of change of quantity.
What is the rate of change of distance with respect to time called?
The rate of change of distance with respect to time is called speed, which is represented as ds/dt.
How to calculate the rate of change?
To calculate the rate of change of quantities, we need to find the derivative of a quantity with respect to another quantity.
What is the example of the rate of change of quantities?
The example of rate of change of quantities is rate of change of speed with respect to time.