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Unit 3
5–10% of examThis unit finishes your derivative toolkit. The chain rule handles a function inside another function, implicit differentiation works when y isn't solved for, and you'll find derivatives of inverse functions, including inverse trig. You'll also take derivatives of derivatives.
Longer videos that cover the whole unit. Good for a first pass or a final review.
If y = f(g(x)), then y′ = f′(g(x))·g′(x): take the derivative of the outer function, leave the inside alone, then multiply by the derivative of the inside. It shows up almost everywhere, including in problems with tables or graphs of f and g.
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When an equation mixes x and y, like x² + y² = 25, differentiate both sides with respect to x and treat y as a function of x, so every y term picks up a dy/dx. Then solve for dy/dx.
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If f(a) = b, then the slope of the inverse function f⁻¹ at x = b is 1/f′(a), as long as f′(a) isn't 0. In practice: find the matching point on f, take the slope of f there, and flip it.
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The inverse trig functions have their own derivative formulas, such as arcsin x → 1/√(1 − x²) and arctan x → 1/(1 + x²). They come from the inverse function idea and are often combined with the chain rule.
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Many problems need several rules at once, like a chain rule inside a product rule. Sometimes rewriting first, such as turning a quotient into a product with a negative exponent, makes the derivative easier.
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The second derivative f″(x) is the derivative of f′(x), and you can keep going to the third derivative and beyond. It is written f″(x), y″ or d²y/dx²; for an implicit equation, differentiate dy/dx again.
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