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Application of derivatives

Sandeep Kumar Prasad
3 hrs ago 0 0
Applications of Derivatives — All Concepts (Class 12)

Applications of Derivatives (AOD) uses derivatives to study the behaviour of functions—whether a quantity is increasing or decreasing, where it reaches maximum or minimum values, how curves behave, and how to solve optimization problems.


1. Rate of Change of Quantities

If y=f(x), then

dxdy

represents the rate of change of y with respect to x.

Example

If

A=πr2

then

drdA=2πr

This represents the rate of change of the area of a circle with respect to its radius.

Chain Rule in Rate of Change

If both quantities depend on time t:

y=f(x),x=g(t)

then

dtdy=dxdy⋅dtdx

Important applications

  • Rate of change of area

  • Rate of change of volume

  • Speed and acceleration

  • Growth and decay

  • Expanding circle or sphere

Example: Area of a circle

A=πr2

If

dtdr=3 cm/s

find the rate of change of area when r=5.

dtdA=2πrdtdr=2π(5)(3)dtdA=30π cm2/s

2. Increasing and Decreasing Functions

Let f(x) be differentiable in an interval.

Increasing Function

If

f′(x)>0

then f(x) is increasing in that interval.

Decreasing Function

If

f′(x)<0

then f(x) is decreasing in that interval.

Constant Function

If

f′(x)=0

throughout an interval, then the function is constant there.


Steps to Find Increasing/Decreasing Intervals

Step 1

Find

f′(x)

Step 2

Find critical points by solving

f′(x)=0

Also consider points where f′(x) does not exist.

Step 3

Divide the domain into intervals.

Step 4

Check the sign of f′(x).

Sign of f′(x) Function
+ Increasing
Decreasing

Example

f(x)=x3−3x

Differentiate:

f′(x)=3x2−3=3(x2−1)=3(x−1)(x+1)

Critical points:

x=−1,x=1

Now check signs:

Interval Sign of f′(x)
(−∞,−1) +
(−1,1)
(1,∞) +

Therefore:

f(x) is increasing on (−∞,−1)∪(1,∞)f(x) is decreasing on (−1,1)

3. Critical Points

A value x=c is called a critical point if:

f′(c)=0

or

f′(c) does not exist

provided f(c) exists.

Critical points are important because maximum and minimum values usually occur at such points.


4. Local Maximum and Local Minimum

Local Maximum

A function has a local maximum at x=a if

f(a)>f(x)

for all nearby values of x.

Graphically:

↗Maximum↘

The function changes from:

Increasing→Decreasing

Therefore:

f′(x):+→−

Local Minimum

A function has a local minimum at x=a if

f(a)<f(x)

for all nearby values of x.

Graphically:

↘Minimum↗

The function changes from:

Decreasing→Increasing

Therefore:

f′(x):−→+

5. First Derivative Test

Suppose f′(a)=0.

Check the sign of f′(x) on both sides of a.

Case 1

+→−

Then:

Local Maximum at x=a

Case 2

−→+

Then:

Local Minimum at x=a

Case 3

+→+

No maximum or minimum.

Case 4

−→−

No maximum or minimum.

Summary

Sign change of f′(x) Result
+→− Local Maximum
−→+ Local Minimum
+→+ Neither
−→− Neither

6. Second Derivative Test

Suppose

f′(a)=0

Then calculate:

f′′(a)

If

f′′(a)<0

then f(a) is a local maximum.

If

f′′(a)>0

then f(a) is a local minimum.

If

f′′(a)=0

the test fails, and we must use another method.


Example

Find the maximum and minimum values of

f(x)=x3−3x

Step 1: First derivative

f′(x)=3x2−33x2−3=0x2=1x=±1

Step 2: Second derivative

f′′(x)=6x

At x=−1:

f′′(−1)=−6<0

Therefore, local maximum.

f(−1)=(−1)3−3(−1)=2Local maximum=2

At x=1:

f′′(1)=6>0

Therefore, local minimum.

f(1)=1−3=−2Local minimum=−2

7. Absolute Maximum and Minimum

These are also called absolute extrema.

For a function f(x) on a closed interval [a,b]:

Steps

  1. Find f′(x).

  2. Find critical points inside (a,b).

  3. Calculate f(x) at:

    • Critical points

    • x=a

    • x=b

  4. Compare all values.

Largest value

Absolute Maximum

Smallest value

Absolute Minimum
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