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Calculus - Derivatives of Polynomial and Trigonometric Functions

Grade 11ICSE

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

🔑Concepts

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The Derivative as a Slope: The derivative f′(x)f'(x) represents the instantaneous rate of change or the slope of the tangent line to a curve at any given point (x,y)(x, y). Visually, imagine a secant line passing through two points on a curve; as the distance between these points hh approaches zero, the secant line rotates to become a tangent line at a single point, and its slope becomes the derivative.

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First Principle of Derivatives: This is the fundamental definition used to find the derivative of a function. It is expressed as f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. Visually, this represents the limit of the ratio of the change in the vertical 'rise' to the change in the horizontal 'run' as the 'run' becomes infinitesimally small.

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Power Rule for Polynomials: This rule simplifies the process of finding derivatives for terms like xnx^n. When you differentiate xnx^n, the exponent nn 'drops down' to become a multiplier in front, and the new power is reduced by exactly 1. Visually, this means higher-degree polynomials (like parabolas) reduce to lower-degree functions (like lines) after differentiation.

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Linearity of Derivatives: The derivative of a sum or difference of functions is equal to the sum or difference of their individual derivatives, such that ddx[f(x)±g(x)]=f′(x)±g′(x)\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x). Additionally, a constant factor kk can be pulled out of the derivative. Visually, if you stretch a graph vertically by a factor of kk, the slope at every point also stretches by that same factor kk.

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Derivatives of Basic Trigonometric Functions: The derivative of sin⁡x\sin x is cos⁡x\cos x, and the derivative of cos⁡x\cos x is −sin⁡x-\sin x. Visually, the slope of the sine wave at any point is exactly the value of the cosine wave at that same xx-coordinate. For instance, at x=0x=0, the sine curve is steepest (slope 1), matching the value cos⁡(0)=1\cos(0) = 1.

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Continuity and Differentiability: For a derivative to exist at a point, the function must be continuous (no breaks) and smooth (no sharp corners). Visually, a graph with a sharp 'V' shape (like an absolute value graph) is not differentiable at the vertex because the tangent line's direction would change abruptly, making the slope undefined at that exact point.

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Physical Interpretation as Rate of Change: In physics, if a function s(t)s(t) represents displacement over time tt, its derivative v(t)=s′(t)v(t) = s'(t) represents instantaneous velocity. Visually, if a distance-time graph is a curve, a steeper slope at a specific time indicates a higher velocity at that moment.

📐Formulae

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

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

ddx(k⋅f(x))=k⋅f′(x)\frac{d}{dx}(k \cdot f(x)) = k \cdot f'(x)

ddx(u±v)=dudx±dvdx\frac{d}{dx}(u \pm v) = \frac{du}{dx} \pm \frac{dv}{dx}

ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x

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

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

ddx(sec⁡x)=sec⁡xtan⁡x\frac{d}{dx}(\sec x) = \sec x \tan x

ddx(cot⁡x)=−csc⁡2x\frac{d}{dx}(\cot x) = -\csc^2 x

ddx(csc⁡x)=−csc⁡xcot⁡x\frac{d}{dx}(\csc x) = -\csc x \cot x

💡Examples

Problem 1:

Find the derivative of the polynomial function f(x)=4x5−3x2+7x−12f(x) = 4x^5 - 3x^2 + 7x - 12.

Solution:

Step 1: Apply the sum and difference rule to differentiate each term separately. ddx(4x5)−ddx(3x2)+ddx(7x)−ddx(12)\frac{d}{dx}(4x^5) - \frac{d}{dx}(3x^2) + \frac{d}{dx}(7x) - \frac{d}{dx}(12) Step 2: Use the power rule ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1} and constant rule. Term 1: 4⋅(5x5−1)=20x44 \cdot (5x^{5-1}) = 20x^4 Term 2: 3⋅(2x2−1)=6x1=6x3 \cdot (2x^{2-1}) = 6x^1 = 6x Term 3: 7⋅(1x1−1)=7⋅x0=7⋅1=77 \cdot (1x^{1-1}) = 7 \cdot x^0 = 7 \cdot 1 = 7 Term 4: The derivative of the constant 12 is 0. Step 3: Combine the results. f′(x)=20x4−6x+7f'(x) = 20x^4 - 6x + 7

Explanation:

We use the power rule for each power of xx. The constant multiple stays in front and multiplies the result of the power rule differentiation. The constant term disappears because its rate of change is zero.

Problem 2:

Differentiate the function y=2sin⁡x+5cos⁡xy = 2\sin x + 5\cos x with respect to xx.

Solution:

Step 1: Identify the trigonometric derivatives needed: ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x and ddx(cos⁡x)=−sin⁡x\frac{d}{dx}(\cos x) = -\sin x. Step 2: Apply the linearity rule to differentiate each trigonometric term. dydx=ddx(2sin⁡x)+ddx(5cos⁡x)\frac{dy}{dx} = \frac{d}{dx}(2\sin x) + \frac{d}{dx}(5\cos x) Step 3: Factor out the constants and substitute the derivatives. dydx=2(cos⁡x)+5(−sin⁡x)\frac{dy}{dx} = 2(\cos x) + 5(-\sin x) Step 4: Simplify the expression. dydx=2cos⁡x−5sin⁡x\frac{dy}{dx} = 2\cos x - 5\sin x

Explanation:

The derivatives of sine and cosine are cyclic but involve a sign change for the cosine derivative. The constants 2 and 5 are preserved as multipliers due to the constant multiple rule.