Maharashtra Board Class 12 Maths Commerce Part I Chapter 4 Applications of Derivatives PDF Download

Applications Of Derivatives

Let's Study

Meaning of Derivatives

Increasing and Decreasing Functions.

Maxima and Minima

Application of derivatives to Economics.

Introduction

Derivatives have a wide range of applications in everyday life. In this chapter, we shall discuss geometrical and physical significance of derivatives and some of their applications such as equation of tangent and normal at a point on the curve, rate measure in physical field, approximate values of functions and extreme values of a function.

Teacher's Note

Derivatives help us find the slope of a line touching a curve at any point. Just like the speed of a car at any instant tells us how fast it is going, the derivative tells us how fast a function is changing.

Exam Trick

Remember: Derivative = Rate of Change = Slope of the tangent line. If the derivative is positive, the function is going up. If negative, it is going down.

Points to Remember

The derivative of a function at a point is the slope of the tangent at that point.
A tangent is a line that touches the curve at exactly one point.
A normal is a line perpendicular to the tangent at the same point.
The slope of normal = negative reciprocal of slope of tangent.

4.1 Meaning Of Derivative:

Let y = f(x) be a continuous function of x. It represents a curve in XY plane.

Let P(a, f(a)) and Q(a + h, f(a + h)) be two points on the curve. Join the points P and Q.

The slope of the chord PQ = \(\frac{f(a+h) - f(a)}{h}\)

Let the point Q move along the curve such that Q→P. Then the secant PQ approaches the tangent at P as h→0

\(\lim_{h \to 0}\) (slope of secant PQ) = \(\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\)

Slope of tangent at P = f'(a) (if limit exists)

Thus, the derivative of a function y = f(x) at any point P(a,b) is the slope of the tangent at the point P(a,b) on the curve.

The slope of the tangent at any point P(a,b) is also called gradient of the curve y = f(x) at point P and is denoted by f'(a) or \(\left(\frac{dy}{dx}\right)_P\).

Normal is a line perpendicular to tangent, passing through the point of tangency.

Therefore, Slope of the normal is the negative reciprocal of slope of tangent.

Thus, slope of normal = \(\frac{-1}{f'(a)}\) = \(\left(\frac{-1}{dy/dx}\right)_P\)

Hence,

(i) The equation of tangent to the curve y = f(x) at the point P(a,b) is given by (y - b) = f'(a)(x - a)

(ii) The equation of normal to the curve y = f(x) at the point P(a,b) is given by (y - b) = \(\frac{-1}{f'(a)}\)(x - a)

Teacher's Note

In real life, if you draw a tangent to a curved road, that tangent shows the direction your car is moving at that exact moment. The slope of that tangent tells how steep the road is at that point.

Exam Trick

Remember the tangent formula: (y - y₁) = m(x - x₁). For normal, just use the negative reciprocal of m. Many students forget the negative sign—do not make this mistake!

Points to Remember

Derivative means the slope of the tangent line at a point.
Tangent equation: (y - b) = f'(a)(x - a)
Normal equation: (y - b) = -1/f'(a) × (x - a)
Normal is always perpendicular to tangent.
Both tangent and normal pass through the same point on the curve.

Solved Examples

1) Find the equation of tangent and normal to the curve y = x² - x + 1 at P(1, 2).

Solution: Given equation of curve is y = x² - x + 1

Differentiating with respect to x

\(\frac{dy}{dx}\) = 2x - 1

\(\left(\frac{dy}{dx}\right)_{P(1,2)}\) = 2(1) - 1 = 2 - 1 = 1

The slope of tangent at P(1, 2) is 1

The equation of tangent is (y - 2) = 1(x - 1)

y - 2 = x - 1

x - y + 1 = 0

Now, The slope of Normal at P(1, 2) is \(\frac{-1}{1}\) = -1

The equation of normal is (y - 2) = -1(x - 1)

y - 2 = -x + 1

x + y - 3 = 0

2) Find the equation of tangent and normal to the curve y = 6 - x² where the normal is parallel to the line x - 4y + 3 = 0.

Solution: Let P(x₁,y₁) be the point on the curve y = 6 - x² where the normal is parallel to the line x - 4y + 3 = 0

Consider, y = 6 - x²

\(\frac{dy}{dx}\) = -2x

\(\left(\frac{dy}{dx}\right)_{x=x_1}\) = -2x₁

The slope of the tangent at P(x₁,y₁) = -2x₁

The slope of the normal at P(x₁,y₁) = \(\frac{1}{2x_1}\)

Now, slope of x - 4y + 3 = 0 is \(\frac{1}{4}\)

The slope of the normal = \(\frac{1}{4}\) (since normal is parallel to given line)

\(\frac{1}{2x_1}\) = \(\frac{1}{4}\)

x₁ = 2

P(x₁,y₁) lies on the curve y = 6 - x²

y₁ = 6 - x₁²

y₁ = 6 - 4

y₁ = 2

The point on the curve is (2, 2)

The slope of tangent at (2,2) is -2x₁ = -2(2) = -4

The equation of tangent is (y - 2) = -4(x - 2)

y - 2 = -4x + 8

4x + y - 10 = 0

The equation of normal is (y - 2) = \(\frac{1}{4}\)(x - 2)

4(y - 2) = 1(x - 2)

4y - 8 = x - 2

x - 4y + 6 = 0

Teacher's Note

When a problem says the tangent or normal is parallel to a line, first find the slope of that line. Then use this slope to find the point on the curve where this condition is satisfied.

Exam Trick

If tangent is parallel to a line, both have the same slope. If normal is parallel to a line, use the fact that normal slope = negative reciprocal of tangent slope. Write the slope of the given line in the form y = mx + c.

Points to Remember

Parallel lines have equal slopes.
Normal slope = -1 / tangent slope.
Always substitute the point back into the curve equation to verify it lies on the curve.
Write the final answer in the form ax + by + c = 0.
Check your arithmetic when calculating slopes and points.

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