मराठी
Tamil Nadu Board of Secondary EducationHSC Arts Class 11
पाठ्यपुस्तक उत्तरे६६९१
संकल्पना स्पष्टीकरण५५
अभ्यासक्रम

Concept of Calculus - Differential Calculus

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Topics

  • Introduction
  • Definition: Differentiation
  • Characteristics
  • Process: Finding the Derivative
  • Significance
  • Derivatives
  • Example
Maharashtra State Board: Class 11

Introduction

  • Differential Calculus is the study of how functions change.
  • It deals with the rate of change of a dependent variable (y) with respect to an independent variable (x).
  • In a function y = f(x), x is the independent variable, and y is the dependent variable.
  • A key application is finding a particle's velocity (y) given its position (x).
Maharashtra State Board: Class 11

Definition: Differentiation

"dy/dx is called the derivative of y with respect to x (which is the rate of change of y with respect to change in x) and the process of finding the derivative is called differentiation."

Maharashtra State Board: Class 11

Characteristics

Feature Description Example
Independent Variable (x) A variable whose value can be chosen freely. Position of a particle (x)
Dependent Variable (y) A variable whose value depends on x; y = f(x). Velocity of a particle (y)
Average Rate of Change The slope of the straight line joining two points A and B on the curve. tan θ = \[\frac {Δy}{Δx}\]
Instantaneous Rate of Change / Derivative The limit of \[\frac {Δy}{Δx}\] as Δx goes to zero (lim Δx → 0). \[\frac {dy}{dx}\] at x = x₀
Geometric Interpretation dy⁄dx at x = x₀ is the slope of the tangent to the curve at that point. Line PQ in Fig. 2.12 (b)
Maharashtra State Board: Class 11

Process: Finding the Derivative

The process of finding the derivative, called differentiation, uses the limit definition:

Average rate of change of y with respect to x.

Rate of change of y with respect to x at x0.

  1. Start with the function: y = f(x).
  2. Define a small change: Take a small increment in x as Δx (from x to x + Δx).
  3. Find the corresponding change in y: The new value of y is f(x + Δx), so Δy = f(x + Δx) − f(x).
  4. Find the ratio (average slope): Calculate \[\frac {Δx}{Δy}\] = \[\frac {f(x + Δx) − f(x)}{Δx}\].
  5. Take the limit: Let Δx approach zero to find the instantaneous slope (the derivative).
    \[\frac{dy}{dx}=\lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\].
Maharashtra State Board: Class 11

Significance

  • \[\frac {dx}{dy}\] is the rate of change of y with respect to the change in x.
  • The derivative at a point is the slope of the tangent to the curve at that point.
  • The properties listed are needed in later chapters (of the physics text).
Maharashtra State Board: Class 11

Derivatives

Name Formula
Derivative (Definition) \[\frac{dy}{dx}=\lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\]
Constant Multiple Rule (s is a constant) \[\frac{d}{dx}(s\cdot f(x))=s\cdot\frac{df(x)}{dx}\]
Addition/Subtraction Rule \[\frac{d}{dx}(f_1(x)+f_2(x))=\frac{df_1(x)}{dx}+\frac{df_2(x)}{dx}\]
Product Rule \[\begin{aligned}
 & \frac{d}{dx}(f_1(x)\times f_2(x))=f_1(x)\frac{df_2(x)}{dx}+ \\
 & f_2(x)\frac{df_1(x)}{dx}
\end{aligned}\]
Quotient Rule \[\frac{d}{dx}\left(\frac{f_1(x)}{f_2(x)}\right)=\frac{1}{f_2(x)}\frac{df_1(x)}{dx}-\frac{f_1(x)}{f_2^2(x)}\frac{df_2(x)}{dx}\]
Chain Rule (Time Dependence) If x depends on t: \[\frac{df(x)}{dt}=\frac{df(x)}{dx}\cdot\frac{dx}{dt}\]
Chain Rule (Function Composition) \[\frac{dy}{dx}=\frac{dy}{d\nu}\cdot\frac{d\nu}{dx}\] (where v is an intermediate variable)
Power Rule \[\frac {d}{dx}\] (xⁿ) = n·xⁿ⁻¹
Exponential Rule (Base e) \[\frac {d}{dx}\] (eˣ) = eˣ and \[\frac {d}{dx}\] (eᵃˣ) = a·eᵃˣ
Natural Log Rule \[\frac {d}{dx}\] (ln x) = 1/x
Sine Rule \[\frac {d}{dx}\] (sin x) = cos x
Cosine Rule \[\frac {d}{dx}\] (cos x) = −sin x
Tangent Rule \[\frac {d}{dx}\](tan x) = sec² x
Cotangent Rule \[\frac {d}{dx}\] (cot x) = −cosec² x
Secant Rule \[\frac {d}{dx}\] (sec x) = tan x · sec x
Cosecant Rule \[\frac {d}{dx}\] (cosec x) = −cosec x · cot x
Maharashtra State Board: Class 11

Example

Problem: Find the derivatives of the functions:

  1. f(x) = x⁸
  2. f(x) = x³ + sinx
  3. f(x) = x³ sinx

Solution:

  1. For f(x) = x⁸
    Rule Used: Power Rule: \[\frac {dx^n}{dx}\] = nxn−1
    Calculation: \[\frac{d(x^{8})}{dx}\] = 8x8−1 = 8x7
  2. For f(x) = x³ + sin x
    Rules Used: Addition Rule (2.27), Power Rule (2.31), Sine Rule (2.34)
    Calculation: \[\frac{d}{dx}(x^3+\sin x)=\frac{d(x^3)}{dx}+\frac{d(\sin x)}{dx}\]
    = 3x2 + cos x
  3. For f(x) = x3 sin x
    Rules Used: Product Rule (2.28), Power Rule (2.31), Sine Rule (2.34)
    Calculation: \[\frac{d}{dx}(x^3\sin x)=x^3\frac{d(\sin x)}{dx}+\frac{d(x^3)}{dx}\sin x\]
    = x³(cos x) + (3x²)sin x
    = x³cos x + 3 x²sin x

Test Yourself

Related QuestionsVIEW ALL [15]

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  1. What does Qt= f (Pt, Y, A, N, Pc) mean in words?
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  2. ₹20,
  3. ₹40.

What is the formula for the elasticity of supply if you know the supply function?

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