In applications of derivatives class 12 chapter 6, we will study different applications of derivatives in various fields like Science, Engineering, and many other fields. In chapter 6, we are going to learn how to determine the rate of change of quantity, finding the equations of tangents, finding turning points on the graphs for various functions, maxima and minima and so on. Let us discuss some important concepts involved in the application of derivatives class 12 in detail.
Application of Derivatives Class 12 Concepts
The topic and subtopics covered in applications of derivatives class 12 chapter 6 are:
- Introduction
- Rate of Change of Quantities
- Increasing and Decreasing Functions
- Tangents and Normals
- Approximations
- Maxima and Minima
- Maximum and Minimum Values of a Function in a Closed Interval
Application of Derivatives Class 12 Notes
Let us discuss the important concepts involved in applications of derivatives class 12 with examples.
Rate of change of quantity-
Consider a function y = f(x), the rate of change of a function is defined as-
dy/dx = f'(x)
Further, if two variables x and y are varying to another variable, say if x = f(t), and y = g(t), then using Chain Rule, we have:
dy/dx = (dy/dt)/(dx/dt)
where dx/dt is not equal to 0.
Increasing and Decreasing Functions:
Consider a function f, continuous in [a,b] and differentiable on the open interval (a,b), then
(i) f is increasing in [a,b] if f'(x)>0 for each x in (a,b)
(ii) f is decreasing in [a,b] if f'(x)< 0 for each x in (a,b)
(iii) f is constant function in [a,b], if f'(x) = 0 for each x in (a,b)
Video Lessons
Tangents & Normals
Rolle’s Theorem & LMVT Theorem
Increasing & Decreasing Functions
Applications Of Monotonicity
Maxima and Minima
Applications of Maxima & Minima
Conceptual Problems
Important Questions
Applications of Derivatives Class 12 Example
Example:
The cube volume is increasing at a rate of 9 cubic centimeters/second. Determine how fast is the surface area increasing when the length of an edge is 10 cm.
Solution:
Let,
x = side length
V = Volume
S = Surface area
Therefore, Volume, V = x3 and surface area, S = 6x2
Where “x” is the function of the time “t”
It is given that, dV/dt = 9 cm3/s
Hence, by using the chain rule, we can write it as:
9 = dV/dt = (d/dt)(x3) = (d/dx)(x3) . (dx/dt)
It becomes,
3x2 . (dx/dt)
or
(dx/dt) = 3/x2 ….(1)
Now for surface area:
dS/dt = (d/dt)(6x2) = (d/dx)(6x2). (dx/dt) (Using Chain Rule)
= 12x. (3/x2) = 36/x ….(Using 1)
Hence, when x = 10 cm, dS/dt = 3.6 cm2/s
Stay tuned with BYJU’S – The Learning App for more class 12 Maths concepts also read related articles to learn the topic with ease.
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