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

Grade 11IB

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

🔑Concepts

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The Derivative as a Gradient: The derivative f′(x)f'(x) represents the instantaneous rate of change or the gradient (slope) of the tangent line to a curve at any given point. Visually, if you imagine a curve on a coordinate plane, the derivative at point xx tells you exactly how steep a straight line would be if it just grazed the curve at that specific point.

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The Power Rule: This is the fundamental rule for differentiating polynomials. For any term xnx^n, the derivative is found by bringing the exponent nn down as a multiplier and reducing the power by one to get nxn−1nx^{n-1}. For a constant cc, the derivative is 00, which is visually represented by a horizontal line having zero slope.

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Linearity of Differentiation: Differentiation is a linear operation, meaning the derivative of a sum or difference of functions is simply the sum or difference of their individual derivatives, such as ddx[f(x)+g(x)]=f′(x)+g′(x)\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x). This allows complex equations to be broken down into simpler parts.

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The Product Rule: When two functions are multiplied together, u(x)⋅v(x)u(x) \cdot v(x), their derivative is not just the product of their derivatives. Instead, it follows the pattern u′v+uv′u'v + uv'. Visually, this accounts for how the total area of a rectangle changes when both its width and height are changing simultaneously.

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The Quotient Rule: This rule is used for functions written as fractions u(x)v(x)\frac{u(x)}{v(x)}. The formula is u′v−uv′v2\frac{u'v - uv'}{v^2}. It is vital to maintain the order in the numerator (derivative of the top times the bottom, minus the top times the derivative of the bottom).

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The Chain Rule: Used for composite functions f(g(x))f(g(x)), often described as a 'function within a function'. You differentiate the 'outer' function while leaving the 'inner' function alone, then multiply by the derivative of the 'inner' function. Visually, this is like a gear system where the rate of change of the outer gear depends on the rate of change of the inner gear it is connected to.

📐Formulae

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

Constant Multiple: ddx(c⋅u)=c⋅dudx\frac{d}{dx}(c \cdot u) = c \cdot \frac{du}{dx}

Sum/Difference Rule: ddx(u±v)=dudx±dvdx\frac{d}{dx}(u \pm v) = \frac{du}{dx} \pm \frac{dv}{dx}

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

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

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

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

💡Examples

Problem 1:

Find the derivative of f(x)=(3x2−5)4f(x) = (3x^2 - 5)^4.

Solution:

Let u=3x2−5u = 3x^2 - 5. Then the function becomes f(u)=u4f(u) = u^4. \n1. Differentiate the outer function: ddu(u4)=4u3\frac{d}{du}(u^4) = 4u^3. \n2. Differentiate the inner function: dudx=ddx(3x2−5)=6x\frac{du}{dx} = \frac{d}{dx}(3x^2 - 5) = 6x. \n3. Apply the Chain Rule: f′(x)=4u3⋅6xf'(x) = 4u^3 \cdot 6x. \n4. Substitute uu back: f′(x)=4(3x2−5)3⋅6x=24x(3x2−5)3f'(x) = 4(3x^2 - 5)^3 \cdot 6x = 24x(3x^2 - 5)^3.

Explanation:

This problem uses the Chain Rule because we have an inner polynomial function raised to an outer power of 4.

Problem 2:

Differentiate y=x22x+1y = \frac{x^2}{2x + 1} with respect to xx.

Solution:

Let u=x2u = x^2 and v=2x+1v = 2x + 1. \n1. Find the derivatives: u′=2xu' = 2x and v′=2v' = 2. \n2. Use the Quotient Rule formula: y′=u′v−uv′v2y' = \frac{u'v - uv'}{v^2}. \n3. Substitute the values: y′=(2x)(2x+1)−(x2)(2)(2x+1)2y' = \frac{(2x)(2x + 1) - (x^2)(2)}{(2x + 1)^2}. \n4. Simplify the numerator: y′=4x2+2x−2x2(2x+1)2=2x2+2x(2x+1)2y' = \frac{4x^2 + 2x - 2x^2}{(2x + 1)^2} = \frac{2x^2 + 2x}{(2x + 1)^2}. \n5. Final factorized form: y′=2x(x+1)(2x+1)2y' = \frac{2x(x + 1)}{(2x + 1)^2}.

Explanation:

The Quotient Rule is applied here as the variable xx appears in both the numerator and the denominator.