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Limits and Derivatives - Intuitive Idea of Derivatives

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Derivative as a Rate of Change: The intuitive idea behind a derivative is the measure of how a function f(x)f(x) changes as its input xx changes. Visually, if you have a graph of a function, the derivative at any point represents how 'steep' the graph is at that specific location. For example, in a distance-time graph, the derivative represents the speed at a particular moment.

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Slope of the Tangent Line: Geometrically, the derivative of a function at a point is the slope of the tangent line to the curve at that point. If you imagine zooming into a curve infinitely at a single point, the curve begins to look like a straight line; the slope of this 'limiting' straight line is the derivative.

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From Secant to Tangent: Consider two points PP and QQ on a curve. A line passing through both is called a secant line. As point QQ moves along the curve toward point PP, the horizontal distance between them, denoted as hh, approaches zero. Visually, the secant line rotates until it perfectly aligns with the tangent line at point PP.

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First Principle of Derivatives: This is the formal mathematical definition of a derivative using limits. It calculates the instantaneous rate of change by taking the limit of the average rate of change (the slope of the secant) over an interval [x,x+h][x, x+h] as the interval width hh shrinks to zero.

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Instantaneous Velocity: In physics, if s(t)s(t) represents the position of an object at time tt, the derivative s′(t)s'(t) represents the instantaneous velocity. On a position-versus-time graph, while the average velocity is the slope of a line between two points, the instantaneous velocity is the slope of the curve at one exact point in time.

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Differentiability and Smoothness: For a derivative to exist at a point, the graph must be 'smooth' and continuous. This means there should be no sharp corners (like the 'V' shape in ∣x∣|x| at x=0x=0), no vertical tangents, and no breaks or jumps in the graph. If a graph has a sharp 'kink', the tangent line is not uniquely defined, and the derivative does not exist there.

📐Formulae

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1}

ddx(c)=0 where c is a constant\frac{d}{dx}(c) = 0 \text{ where } c \text{ is a constant}

ddx[f(x)±g(x)]=f′(x)±g′(x)\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)

💡Examples

Problem 1:

Find the derivative of the function f(x)=x2f(x) = x^2 at the point x=3x = 3 using the first principle.

Solution:

Step 1: Identify f(a)f(a) and f(a+h)f(a+h). Here a=3a = 3. f(3)=32=9f(3) = 3^2 = 9 f(3+h)=(3+h)2=9+6h+h2f(3+h) = (3+h)^2 = 9 + 6h + h^2 Step 2: Substitute these into the limit definition formula: f′(3)=lim⁡h→0f(3+h)−f(3)hf'(3) = \lim_{h \to 0} \frac{f(3+h) - f(3)}{h} f′(3)=lim⁡h→0(9+6h+h2)−9hf'(3) = \lim_{h \to 0} \frac{(9 + 6h + h^2) - 9}{h} Step 3: Simplify the expression: f′(3)=lim⁡h→06h+h2h=lim⁡h→0(6+h)f'(3) = \lim_{h \to 0} \frac{6h + h^2}{h} = \lim_{h \to 0} (6 + h) Step 4: Evaluate the limit as h→0h \to 0: f′(3)=6+0=6f'(3) = 6 + 0 = 6 Therefore, the derivative of x2x^2 at x=3x=3 is 66.

Explanation:

We use the definition of the derivative at a specific point. By expanding the binomial (3+h)2(3+h)^2, we are able to cancel out the constant terms, divide by hh to remove the indeterminate form, and then safely apply the limit.

Problem 2:

Find the general derivative of f(x)=5x+2f(x) = 5x + 2 using the intuitive limit method.

Solution:

Step 1: Write the expressions for f(x)f(x) and f(x+h)f(x+h): f(x)=5x+2f(x) = 5x + 2 f(x+h)=5(x+h)+2=5x+5h+2f(x+h) = 5(x+h) + 2 = 5x + 5h + 2 Step 2: Apply the first principle formula: f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} f′(x)=lim⁡h→0(5x+5h+2)−(5x+2)hf'(x) = \lim_{h \to 0} \frac{(5x + 5h + 2) - (5x + 2)}{h} Step 3: Simplify the numerator: f′(x)=lim⁡h→05hhf'(x) = \lim_{h \to 0} \frac{5h}{h} Step 4: Evaluate the limit: f′(x)=lim⁡h→05=5f'(x) = \lim_{h \to 0} 5 = 5 The derivative is 55.

Explanation:

Since f(x)=5x+2f(x) = 5x + 2 is a linear function (a straight line), its slope is constant everywhere. The derivative represents this slope. The limit process confirms that the rate of change is consistently 55 regardless of the value of xx.