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Limits and Derivatives - Derivatives

Grade 11CBSE

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

🔑Concepts

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Derivative at a Point: The derivative of a function ff at a point aa in its domain is the instantaneous rate of change of the function at that point. Mathematically, it is defined as f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}, provided the limit exists.

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Geometric Interpretation: The derivative f′(x)f'(x) at any point PP on a curve y=f(x)y = f(x) represents the slope of the tangent line to the curve at that point. Visually, as we zoom into a point on a smooth curve, the curve starts to look like a straight line; the slope of this 'local' line is the derivative.

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First Principle of Derivative: This is the method of finding the derivative of a function using the basic limit definition. It involves calculating the limit of the ratio of the change in the function to the change in the independent variable as the change approaches zero.

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Physical Meaning of Derivative: In physics, if a function represents the position of an object with respect to time (s=f(t)s = f(t)), the derivative f′(t)f'(t) represents the instantaneous velocity. Visually, this is the steepness of the position-time graph at any given moment.

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Algebra of Derivatives: Derivatives follow specific algebraic properties: the derivative of a sum or difference of two functions is the sum or difference of their derivatives, and the derivative of a constant multiplied by a function is the constant times the derivative of the function.

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Product Rule (Leibniz Rule): When differentiating the product of two functions u(x)u(x) and v(x)v(x), the derivative is not simply the product of their derivatives. Instead, it is the first function times the derivative of the second, plus the second function times the derivative of the first.

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Quotient Rule: For a function in the form of a fraction u(x)v(x)\frac{u(x)}{v(x)}, where v(x)≠0v(x) \neq 0, the derivative is calculated by taking the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

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Differentiability and Continuity: For a function to have a derivative at a point, it must be continuous at that point. Visually, a function is not differentiable at 'sharp corners' (cusps) or 'breaks' (discontinuities) in the graph, even if the function is defined there.

📐Formulae

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

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

Constant Rule: ddx(c)=0\frac{d}{dx}(c) = 0, where cc is a constant

Sum/Difference Rule: ddx[f(x)±g(x)]=f′(x)±g′(x)\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)

Product Rule: ddx[u(x)⋅v(x)]=u(x)v′(x)+v(x)u′(x)\frac{d}{dx}[u(x) \cdot v(x)] = u(x)v'(x) + v(x)u'(x)

Quotient Rule: ddx[u(x)v(x)]=v(x)u′(x)−u(x)v′(x)[v(x)]2\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}

Trigonometric Derivative (Sine): ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x

Trigonometric Derivative (Cosine): ddx(cos⁡x)=−sin⁡x\frac{d}{dx}(\cos x) = -\sin x

Trigonometric Derivative (Tangent): ddx(tan⁡x)=sec⁡2x\frac{d}{dx}(\tan x) = \sec^2 x

💡Examples

Problem 1:

Find the derivative of f(x)=x2+3xf(x) = x^2 + 3x using the First Principle.

Solution:

Step 1: Write the definition of the derivative: f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} Step 2: Substitute f(x+h)f(x+h) and f(x)f(x) into the formula: f′(x)=lim⁡h→0[(x+h)2+3(x+h)]−[x2+3x]hf'(x) = \lim_{h \to 0} \frac{[(x+h)^2 + 3(x+h)] - [x^2 + 3x]}{h} Step 3: Expand the terms: f′(x)=lim⁡h→0x2+2xh+h2+3x+3h−x2−3xhf'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 + 3x + 3h - x^2 - 3x}{h} Step 4: Simplify the numerator: f′(x)=lim⁡h→02xh+h2+3hhf'(x) = \lim_{h \to 0} \frac{2xh + h^2 + 3h}{h} Step 5: Factor out hh and cancel: f′(x)=lim⁡h→0(2x+h+3)f'(x) = \lim_{h \to 0} (2x + h + 3) Step 6: Apply the limit h→0h \to 0: f′(x)=2x+0+3=2x+3f'(x) = 2x + 0 + 3 = 2x + 3

Explanation:

We use the limit definition to find the rate of change. By expanding (x+h)2(x+h)^2, we identify the terms that change relative to hh and eliminate the original function terms (x2x^2 and 3x3x).

Problem 2:

Differentiate y=xsin⁡xy = x \sin x with respect to xx.

Solution:

Step 1: Identify the functions for the product rule: let u(x)=xu(x) = x and v(x)=sin⁡xv(x) = \sin x. Step 2: Find their individual derivatives: u′(x)=ddx(x)=1u'(x) = \frac{d}{dx}(x) = 1 and v′(x)=ddx(sin⁡x)=cos⁡xv'(x) = \frac{d}{dx}(\sin x) = \cos x. Step 3: Apply the product rule formula: dydx=u(x)v′(x)+v(x)u′(x)\frac{dy}{dx} = u(x)v'(x) + v(x)u'(x). Step 4: Substitute the values: dydx=(x)(cos⁡x)+(sin⁡x)(1)\frac{dy}{dx} = (x)(\cos x) + (\sin x)(1). Step 5: Simplify: dydx=xcos⁡x+sin⁡x\frac{dy}{dx} = x \cos x + \sin x.

Explanation:

Since the function is a product of an algebraic term (xx) and a trigonometric term (sin⁡x\sin x), we apply the Product Rule (Leibniz Rule) to find the total derivative.