differentiation rules and techniques — HSC mathematics advanced, live

Differentiation rules and techniques

The first job with any derivative is deciding which rule it needs, and the outermost operation decides. Two functions multiplied need the product rule, u′v + uv′. One divided by another needs the quotient rule, (vu′ − uv′)/v². A function inside another needs the chain rule, the outside derivative times the inside derivative. Some functions need two rules at once: name the outermost one first, then work inwards. This is the real knowscape from knowhere, not a picture of one. Drag it. Watch what actually changes.

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why this one carries the topic.

Calculus is one machine run in two directions: differentiation takes a function and hands you its rate of change; integration takes the rate and hands the function back. The Fundamental Theorem is the proof they're inverses — which is why every substitution is just differentiation read backwards.

Pick a function and decide first whether it needs the product, quotient or chain rule: the working and the bars for each piece of the gradient only open once you name the right rule. Then move a point along the curve and a tiny chord confirms the tangent gradient.

what examiners catch — Assessment tasks check that you differentiate the inside as well as the outside. For (3x² + 1)⁵ the slip is writing 5(3x² + 1)⁴ and stopping; the full answer multiplies by 6x. In the quotient rule, keep vu′ first: swapping the order flips the sign. Write u, v, u′ and v′ down before combining.

five things, not forty.

Product rule: if y = uv then dy/dx = u′v + uv′, so x²(3x + 1) differentiates to 2x(3x + 1) + 3x² = 9x² + 2x.

Quotient rule: if y = u/v then dy/dx = (vu′ − uv′)/v², so (2x + 1)/(x² + 1) gives (2 − 2x − 2x²)/(x² + 1)².

Chain rule: if y = f(g(x)) then dy/dx = f′(g(x)) × g′(x), so (3x² + 1)⁵ gives 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴.

To identify the rule, find the outermost operation: a product, a quotient or a function of a function.

Combinations are common: x(x² − 1)² needs the product rule, and its second factor needs the chain rule to find v′.

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this conceptdifferentiation rules and techniquesPick a function and decide first whether it needs the product, quotient or chain rule: the working and the bars for each piece of the gradient only open once you name the right rule.

sits under itthe derivative as an instantaneous rate of changebecause the rules are shortcuts to a thing you must first understand

built on itimplicit differentiation and related ratesbecause implicit differentiation is the chain rule applied to every y-term

built on itthe chain, product and quotient rulesbecause the advanced rules extend the basic kit

built on itstationary points and their naturebecause finding f′=0 requires finding f′

built on ittangent lines and linear approximationbecause the tangent's slope is the derivative, evaluated

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antiderivatives and indefinite integrationlive →definite integration and the fundamental theorem of calculusin the appintegration by recognition and substitutionin the app

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