# Applications of Derivatives Class 12 Notes Maths Chapter 6 - CBSE

## What are Applications of Derivatives ?

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## Rate Of Change Of Bodies

If a quantity y varies with another quantity x, satisfying some rule y = f(x), then dy/dx represents the rate of change of y w.r.t x and dy/dx x=x0 represents the rate of change of y w.r.t x at x=x0.

## Increasing/decreasing Functions

A function f is said to be :

• Increasing on an interval (a,b) if x1 < x2 in (a,b) ⇒ f (x1) ≤ f (x2) for all x1, x2 (a,b).
• Decreasing on an interval (a,b) if x1 < x2 in (a,b) ⇒ f (x1) ≥ f (x2) for all x1, x2 (a,b).

## Maxima And Minima

First Derivative Test

Let f be a function defined on an open interval I. Let f be continuous at a critical point a in I. Then

• If f '(x) > 0 at every point sufficiently close to and to the left of a, and f '(x) < 0 at every point sufficiently close to and to the right of a, then a is a point of local maxima.
• If f '(x) < 0 at every point sufficiently close to and to the left of a, and f '(x) > 0 at every point sufficiently close to and to the right of a, then a is a point of local minima.
• If f '(x) does not change sign as x increases through a, then a is neither a point of local maxima nor a point of local minima. Infact, such a point is called point of inflexion.

Second Derivative Test

Let f be a function defined on an interval I and a I. Let f be twice differential at c. Then

• x = a is a point of local maxima if f '(a) = 0 and f ''(a) < 0 The values f(a) is local maximum value of f.
• x = a is a point of local minima if f '(a) = 0 and f ''(a) > 0 In this case, f(a) is local minimum value of f.
• The test fails if f '(a) = 0 and f ''(a) = 0 In this case, we go back to the first derivative test and find whether a is a point of maxima, minima or a point of inflexion.

## Working rule for finding absolute maxima and/or absolute minima

• Step: 1 Find all critical points of f in the interval, i.e., find points x where either f '(x) = 0 or f is not differentiable.
• Step: 2 Take the end points of the interval.
• Step: 3 At all these points (listed in Step 1 and 2), calculate the values of f.
• Step: 4 Identify the maximum and minimum values if f out of the values calculated in step 3. This maximum value will be the absolute maximum value of f and the minimum value will be the absolute minimum value of f.