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Calculus - Rules of Differentiation (Product, Quotient, Chain)

Grade 12IB

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

🔑Concepts

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The Chain Rule is used for differentiating composite functions f(g(x))f(g(x)), where one function is nested inside another. Visually, this is like an onion where you differentiate the outer layer while keeping the inner layer intact, then multiply by the derivative of the inner layer. On a graph, this represents how the rate of change of the outer function is scaled by the rate of change of the internal input.

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The Product Rule is applied when two functions are multiplied together, such as u(x)⋅v(x)u(x) \cdot v(x). Think of this visually as the rate of change of the area of a rectangle where both the length and width are changing simultaneously. The total change is the sum of the change in one dimension multiplied by the constant length of the other.

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The Quotient Rule is used for functions expressed as a fraction u(x)v(x)\frac{u(x)}{v(x)}. It is important to remember the order of the numerator (vu′−uv′v u' - u v') and that the denominator becomes v2v^2. Visually, the squaring of the denominator reflects how the magnitude of the divisor impacts the sensitivity of the overall ratio.

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Differentiability implies smoothness. For a derivative to exist at a point, the graph must not have any 'sharp' corners, vertical tangents, or breaks. Visually, if you zoom in infinitely on a differentiable point, the curve should eventually look like a straight line (local linearity).

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The derivative at a specific point represents the gradient of the tangent line at that point. Visually, the tangent line is a straight line that 'just touches' the curve at that coordinate, mirroring the slope of the curve at that exact instant.

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Higher-order derivatives, such as the second derivative f′′(x)f''(x), describe the 'rate of change of the rate of change.' Visually, the second derivative tells us about the concavity of the graph: if f′′(x)>0f''(x) > 0, the graph is concave up (shaped like a cup), and if f′′(x)<0f''(x) < 0, it is concave down (shaped like a frown).

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Combining rules is often necessary for complex expressions. For example, a quotient might contain a product in the numerator, or a product might involve a composite function requiring the chain rule. Visually, these problems require a hierarchical approach, breaking the complex expression into nested parts and applying the rules from the outside in.

📐Formulae

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

Product Rule: ddx(u⋅v)=udvdx+vdudx\text{Product Rule: } \frac{d}{dx}(u \cdot v) = u \frac{dv}{dx} + v \frac{du}{dx}

Quotient Rule: ddx(uv)=vdudx−udvdxv2\text{Quotient Rule: } \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}

Chain Rule (Leibniz notation): dydx=dydu⋅dudx\text{Chain Rule (Leibniz notation): } \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

Chain Rule (Function notation): [f(g(x))]′=f′(g(x))⋅g′(x)\text{Chain Rule (Function notation): } [f(g(x))]' = f'(g(x)) \cdot g'(x)

Derivative of ekx:ddx(ekx)=kekx\text{Derivative of } e^{kx}: \frac{d}{dx}(e^{kx}) = ke^{kx}

Derivative of ln⁡(x):ddx(ln⁡(x))=1x\text{Derivative of } \ln(x): \frac{d}{dx}(\ln(x)) = \frac{1}{x}

💡Examples

Problem 1:

Differentiate f(x)=(x3+2x)5f(x) = (x^3 + 2x)^5 with respect to xx.

Solution:

Step 1: Identify the outer function u5u^5 and the inner function u=x3+2xu = x^3 + 2x. \ Step 2: Apply the Chain Rule: f′(x)=5(u)4⋅dudxf'(x) = 5(u)^4 \cdot \frac{du}{dx}. \ Step 3: Differentiate the inner function: dudx=3x2+2\frac{du}{dx} = 3x^2 + 2. \ Step 4: Substitute back: f′(x)=5(x3+2x)4⋅(3x2+2)f'(x) = 5(x^3 + 2x)^4 \cdot (3x^2 + 2). \ Final Answer: f′(x)=5(3x2+2)(x3+2x)4f'(x) = 5(3x^2 + 2)(x^3 + 2x)^4.

Explanation:

This problem requires the Chain Rule because we have a polynomial function raised to a power. We treat the parentheses as a single variable first, differentiate the outer power, and then multiply by the derivative of what is inside.

Problem 2:

Find the derivative of y=x2sin⁡(x)y = \frac{x^2}{\sin(x)}.

Solution:

Step 1: Identify u=x2u = x^2 and v=sin⁡(x)v = \sin(x). \ Step 2: Differentiate both: u′=2xu' = 2x and v′=cos⁡(x)v' = \cos(x). \ Step 3: Apply the Quotient Rule: dydx=vu′−uv′v2\frac{dy}{dx} = \frac{v u' - u v'}{v^2}. \ Step 4: Substitute the terms: dydx=(sin⁡(x))(2x)−(x2)(cos⁡(x))(sin⁡(x))2\frac{dy}{dx} = \frac{(\sin(x))(2x) - (x^2)(\cos(x))}{(\sin(x))^2}. \ Final Answer: dydx=2xsin⁡(x)−x2cos⁡(x)sin⁡2(x)\frac{dy}{dx} = \frac{2x\sin(x) - x^2\cos(x)}{\sin^2(x)}.

Explanation:

The function is a ratio of two distinct functions, so the Quotient Rule is mandatory. We carefully identify the numerator and denominator to ensure the subtraction in the numerator of the formula is in the correct order.