krit.club logo

Calculus - Derivatives of Polynomial and Trigonometric Functions

Grade 11ICSE

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

🔑Concepts

•

The derivative of a function at a point xx represents the slope of the tangent line to the curve at that point. For a polynomial curve like y=x2y = x^2, the slope changes continuously along the curve.

Graph of y=x^2 showing a tangent line at x=2 illustrating the derivative as slope.
•

The Power Rule states that for any real number nn, ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1}. This allows for quick differentiation of terms like x3x^3 (becoming 3x23x^2) or x\sqrt{x} (which is x1/2x^{1/2}).

•

Trigonometric functions exhibit periodic rates of change. The derivative of sin⁡x\sin x is cos⁡x\cos x, which means the slope of the sine wave at any point is given by the value of the cosine function at that same point.

Graph comparing sin(x) and its derivative cos(x).
•

Linearity of Differentiation: The derivative of a sum of functions is the sum of their derivatives, and constants can be factored out. ddx[af(x)+bg(x)]=af′(x)+bg′(x)\frac{d}{dx}[af(x) + bg(x)] = a f'(x) + b g'(x)

📐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.

Problem 3:

Differentiate the function y=13x3−4x+10y = \frac{1}{3}x^3 - 4\sqrt{x} + 10 with respect to xx.

Graph of the function y = 1/3x^3 - 4x^0.5 + 10.

Solution:

  1. Rewrite the terms in power form: y=13x3−4x1/2+10y = \frac{1}{3}x^3 - 4x^{1/2} + 10
  2. Apply the power rule ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1} to each term: dydx=13(3x3−1)−4(12x1/2−1)+0\frac{dy}{dx} = \frac{1}{3}(3x^{3-1}) - 4(\frac{1}{2}x^{1/2-1}) + 0
  3. Simplify the coefficients and exponents: dydx=x2−2x−1/2\frac{dy}{dx} = x^2 - 2x^{-1/2}
  4. Final form: dydx=x2−2x\frac{dy}{dx} = x^2 - \frac{2}{\sqrt{x}}

Explanation:

This example demonstrates using the power rule for integer and fractional exponents, along with the constant rule (the derivative of 10 is 0).

Problem 4:

Calculate the slope of the curve f(x)=3tan⁡x−2sin⁡xf(x) = 3\tan x - 2\sin x at the point where x=0x = 0.

Graph of f(x) = 3tan(x) - 2sin(x) near the origin with a tangent line of slope 1.

Solution:

  1. Find the general derivative f′(x)f'(x): f′(x)=ddx(3tan⁡x)−ddx(2sin⁡x)f'(x) = \frac{d}{dx}(3\tan x) - \frac{d}{dx}(2\sin x)
  2. Use trigonometric derivative identities: f′(x)=3sec⁡2x−2cos⁡xf'(x) = 3\sec^2 x - 2\cos x
  3. Substitute x=0x = 0 into the derivative to find the slope: f′(0)=3sec⁡2(0)−2cos⁡(0)f'(0) = 3\sec^2(0) - 2\cos(0)
  4. Evaluate trigonometric values: sec⁡(0)=1\sec(0) = 1 and cos⁡(0)=1\cos(0) = 1 f′(0)=3(1)2−2(1)=3−2=1f'(0) = 3(1)^2 - 2(1) = 3 - 2 = 1 The slope of the curve at x=0x = 0 is 11.

Explanation:

The derivative function provides the slope at any point. By substituting a specific value of xx, we find the instantaneous rate of change at that specific location.