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Algebra and Graphs - Basic differentiation and gradients

Grade 11IGCSE

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

🔑Concepts

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The derivative dy/dx represents the gradient (slope) of the tangent to a curve at any given point.

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Differentiation is the process of finding the derivative function.

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The Power Rule: If y = ax^n, then dy/dx = anx^{n-1}.

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The derivative of a constant is always 0.

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Stationary points (turning points) occur where the gradient of the curve is zero (dy/dx = 0).

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A tangent is a straight line that just touches a curve at a point and has the same gradient as the curve at that point.

📐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)

m=dydxm = \frac{dy}{dx} at x=ax = a

Equation of tangent: y−y1=m(x−x1)y - y_1 = m(x - x_1)

💡Examples

Problem 1:

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

Solution:

dydx=15x2−4x+4\frac{dy}{dx} = 15x^2 - 4x + 4

Explanation:

Apply the power rule to each term: 3×5x(3−1)=15x23 \times 5x^{(3-1)} = 15x^2, 2×2x(2−1)=4x2 \times 2x^{(2-1)} = 4x, the derivative of 4x4x is 44, and the derivative of the constant −7-7 is 00.

Problem 2:

Find the gradient of the curve y=x2+3xy = x^2 + 3x at the point where x=2x = 2.

Solution:

77

Explanation:

First, find the derivative: dydx=2x+3\frac{dy}{dx} = 2x + 3. To find the gradient at the specific point x=2x = 2, substitute 22 into the derivative: 2(2)+3=4+3=72(2) + 3 = 4 + 3 = 7.

Problem 3:

Find the coordinates of the stationary point on the curve y=x2−6x+5y = x^2 - 6x + 5.

Solution:

(3,−4)(3, -4)

Explanation:

  1. Find dydx=2x−6\frac{dy}{dx} = 2x - 6. 2. Set dydx=0\frac{dy}{dx} = 0 for stationary points: 2x−6=0  ⟹  x=32x - 6 = 0 \implies x = 3. 3. Substitute x=3x = 3 back into the original equation to find yy: y=(3)2−6(3)+5=9−18+5=−4y = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4.