 # Application of Derivatives - II

## We can apply derivatives to determine approximate value of certain quantities and to find maximum and minimum values of a function in an interval. The differential of y is denoted by dy and is defined by dy = f ′(x) dx or dy = f ′(x) ∆x, where dy approximately equal to ∆y. First Derivative Test for Maxima and Minima: Let f (x) be a function defined on an open interval I and continuous at a critical point ‘a’ in I. Then If f ′(x) changes sign from positive to negative as x increases through a, i.e., 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) changes sign from negative to positive as x increases through a, i.e., 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. Second Derivative Test  For Maxima and Minima: Let f(x) be a function defined on an interval I and a Î I. Let f(x) be twice differentiable at a. Then x = a is a point of local maxima, and if f ′(a) = 0 and f ″(a) < 0, then f(a) is the local maximum value of f (x). x = a is a point of local minima, and if f′(a) = 0 and f ″(a) > 0, the f (a) is the local minimum value of f (x). The test fails if f'(a) = 0 and f" (a) = 0.        Keywords: Maximum and minimum values of a function, First derivative test, Second derivative test

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• Q1 Marks:1

2.

##### Explanation: • Q2

The maximum and minimum values of 4 cos x + 3 sin x are respectively

Marks:1

5 and -5.

##### Explanation:
The maximum and minimum values of a cos x + b sin x are (a2+b2) and - (a2+b2) respectively.
• Q3

The approximate change in the volume V of a cube of side a metres caused by increasing the side by 5% is (in m3)

Marks:1

0.15 a3.

##### Explanation: • Q4

If f(x) = 3x2 + 5x + 3, then the approximate value of f(3.02) is

Marks:1

45.46.

##### Explanation: • Q5

Approximate value of (0.99) is

Marks:1 