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Calculus - Differentiation of Polynomials

Grade 12IGCSE

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

🔑Concepts

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Definition of the derivative as the gradient function of a curve.

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The Power Rule for differentiation: applicable to positive, negative, and fractional indices.

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Linearity of differentiation: the derivative of a sum is the sum of the derivatives.

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Constant Rule: The derivative of any constant value is always zero.

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Notation: Using dydx\frac{dy}{dx} for Leibniz notation and f′(x)f'(x) for Lagrange notation.

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Applications: Finding the gradient of a tangent at a specific point (x,y)(x, y).

📐Formulae

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

ddx(axn)=anxn−1\frac{d}{dx}(ax^n) = anx^{n-1}

ddx(c)=0\frac{d}{dx}(c) = 0, where cc 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:

Differentiate y=4x3−5x2+7x−10y = 4x^3 - 5x^2 + 7x - 10 with respect to xx.

Solution:

dydx=12x2−10x+7\frac{dy}{dx} = 12x^2 - 10x + 7

Explanation:

Apply the power rule to each term: 3×4x3−1=12x23 \times 4x^{3-1} = 12x^2; 2×−5x2−1=−10x2 \times -5x^{2-1} = -10x; the derivative of 7x7x is 7, and the constant -10 becomes 0.

Problem 2:

Find f′(x)f'(x) if f(x)=2x2+xf(x) = \frac{2}{x^2} + \sqrt{x}.

Solution:

f′(x)=−4x−3+12x−1/2f'(x) = -4x^{-3} + \frac{1}{2}x^{-1/2} or −4x3+12x-\frac{4}{x^3} + \frac{1}{2\sqrt{x}}

Explanation:

First, rewrite the expression in index form: f(x)=2x−2+x1/2f(x) = 2x^{-2} + x^{1/2}. Then apply the power rule: −2×2x−3=−4x−3-2 \times 2x^{-3} = -4x^{-3} and 12x1/2−1=12x−1/2\frac{1}{2}x^{1/2-1} = \frac{1}{2}x^{-1/2}.

Problem 3:

Find the gradient of the curve y=x2−4x+1y = x^2 - 4x + 1 at the point where x=3x = 3.

Solution:

Gradient = 2

Explanation:

First, find the derivative dydx=2x−4\frac{dy}{dx} = 2x - 4. To find the gradient at a specific point, substitute x=3x = 3 into the derivative: 2(3)−4=6−4=22(3) - 4 = 6 - 4 = 2.