Source: MTH 132 Calculus I, Michigan State University, Chapter 2 – Derivatives (Sections 4–6)
Difficulty: Intermediate | Prerequisites: Part 1 study notes (limit definition, power/product/quotient rules), basic trig identities
With the basic differentiation rules from Part 1 in hand, this part adds three major tools. First, the derivatives of the six trigonometric functions. Second, the chain rule, which handles compositions of functions (anything with a function inside another function). Third, implicit differentiation, which lets you find slopes of curves that are not written as y = f(x). Together, these cover nearly every derivative you will see on the exam.
The derivatives of sin and cos are cos and −sin respectively; the other four trig derivatives follow from the quotient rule. The chain rule says "derivative of the outside times derivative of the inside." Implicit differentiation differentiates both sides of an equation with respect to x, treating y as a function of x, then solves for dy/dx.
Special trigonometric limits
lim(x→0) sin(x)/x = 1 and lim(x→0) (cos(x) − 1)/x = 0. These two results are essential for proving the derivatives of sin and cos from the limit definition, and they appear in their own exam problems.
In simple terms, for very small angles (in radians), sin(x) is approximately equal to x.
Composition of functions
f(g(x)), also written (f ∘ g)(x). The output of g becomes the input of f. Recognising a function as a composition is the prerequisite for applying the chain rule.
Think of it as one function "eating" another: if h(x) = sin(x² + x), the outer function is sin and the inner function is x² + x.
Chain rule
If F(x) = f(g(x)), then F'(x) = f'(g(x)) · g'(x). In Leibniz notation with u = g(x): dy/dx = (dy/du) · (du/dx).
In simple terms, differentiate the outside (leaving the inside untouched), then multiply by the derivative of the inside.
Implicit differentiation
A method of differentiating both sides of an equation with respect to x and then solving the resulting equation for y' (or dy/dx). Used when the equation defines y as a function of x implicitly rather than explicitly.
Think of it as: every time you differentiate a term containing y, tack on a factor of dy/dx via the chain rule, because y depends on x.
(d/dx) sin(x) = cos(x)
(d/dx) cos(x) = −sin(x)
The proof of (d/dx) sin(x) = cos(x) uses the limit definition, the angle-addition formula sin(x+h) = sin(x)cos(h) + cos(x)sin(h), and the two special limits.
All derived from sin and cos using the quotient rule:
(d/dx) tan(x) = sec²(x)
(d/dx) sec(x) = sec(x)·tan(x)
(d/dx) csc(x) = −csc(x)·cot(x)
(d/dx) cot(x) = −csc²(x)
Memory aid: the three "co" functions (cos, csc, cot) all pick up a negative sign in their derivatives.
lim(x→0) sin(x)/x = 1 and lim(x→0) (cos(x) − 1)/x = 0.
More generally, if lim(x→a) f(x) = 0, then lim(x→a) sin(f(x))/f(x) = 1 and lim(x→a) f(x)/sin(f(x)) = 1.
These generalisations let you handle limits like lim(x→0) sin(5x)/(3x) = 5/3 by rewriting as (5/3)·[sin(5x)/(5x)] and noting that sin(5x)/(5x) → 1.
Once you know the six trig derivatives, you combine them freely with the product and quotient rules. For example, differentiating p(t) = [4tan(t) − 5]/sec(t) uses the quotient rule with the trig derivative formulas.
f(g(x)) means "apply g first, then f." Before you can use the chain rule, you need to spot the outer and inner functions. Example: h(x) = sin(x² + x) has outer f(u) = sin(u) and inner g(x) = x² + x.
If g is differentiable at x and f is differentiable at g(x), then the composition F = f ∘ g is differentiable at x and F'(x) = f'(g(x)) · g'(x).
In Leibniz notation with u = g(x): dy/dx = (dy/du) · (du/dx). The du terms look like they cancel (they do not, but the visual is a reliable memory aid).
Identify the outer function and the inner function.
Differentiate the outer function, leaving the inner function in place.
Multiply by the derivative of the inner function.
f(x) = (7x³ + 2x² − x + 3)^5. Outer: u^5, inner: 7x³ + 2x² − x + 3. Result: 5(7x³ + 2x² − x + 3)^4 · (21x² + 4x − 1).
g(x) = tan(3x² + 5x + 8). Outer: tan(u), inner: 3x² + 5x + 8. Result: sec²(3x² + 5x + 8) · (6x + 5).
f(x) = cos^6(x) = (cos x)^6. Outer: u^6, inner: cos x. Result: 6(cos x)^5 · (−sin x) = −6cos^5(x)sin(x).
Nested chain rule: f(x) = 3tan(2cos(5x)). Differentiate layer by layer: 3sec²(2cos(5x)) · (−2sin(5x)) · 5.
If h(x) = f(g(x)), then h'(x) = f'(g(x)) · g'(x). When given a table of f, g, f', g' values, just substitute. Example: if g(1) = 2 and f'(2) = 5 and g'(1) = 6, then h'(1) = f'(g(1)) · g'(1) = 5 · 6 = 30.
Some curves cannot be written as y = f(x). For example, x² + y² = 2 is a circle, and x³ + y³ = 6xy (the folium of Descartes) has no neat explicit formula for y. Implicit differentiation lets you find dy/dx without solving for y first.
Differentiate every term on both sides with respect to x.
Whenever you differentiate an expression involving y, apply the chain rule: (d/dx)[stuff with y] = (d/dy)[stuff with y] · dy/dx. In practice, this means every time you differentiate a y term, tack on dy/dx.
Collect all terms with dy/dx on one side.
Factor out dy/dx and solve.
Differentiate both sides: 3x² + 3y²·(dy/dx) = 6y + 6x·(dy/dx).
Rearrange: 3y²·(dy/dx) − 6x·(dy/dx) = 6y − 3x².
Factor: (dy/dx)(3y² − 6x) = 6y − 3x².
Solve: dy/dx = (6y − 3x²)/(3y² − 6x) = (2y − x²)/(y² − 2x).
Tangent line at (3, 3): dy/dx = (6 − 9)/(9 − 6) = −1. Tangent line: y − 3 = −1(x − 3), or y = −x + 6.
Horizontal tangent: set the numerator of dy/dx = 0 (and check the denominator is not also zero).
Vertical tangent: set the denominator of dy/dx = 0 (and check the numerator is not also zero).
You must then use the original equation to find the actual (x, y) point on the curve.
The same method works when trig functions appear. Example: y·sin(2x) = x·cos(2y). Differentiate both sides, apply the product rule to each side, chain-rule the trig arguments, collect dy/dx terms, and solve.
You can find y'' by differentiating dy/dx again with respect to x, substituting dy/dx back in wherever it appears. Example: for √x + √y = 7, first find y' = −√y/√x, then differentiate again (using the quotient rule and chain rule) to get y''.
Formula | Rule |
|---|---|
(d/dx) sin(x) = cos(x) | Sine derivative |
(d/dx) cos(x) = −sin(x) | Cosine derivative |
(d/dx) tan(x) = sec²(x) | Tangent derivative |
(d/dx) sec(x) = sec(x)·tan(x) | Secant derivative |
(d/dx) csc(x) = −csc(x)·cot(x) | Cosecant derivative |
(d/dx) cot(x) = −csc²(x) | Cotangent derivative |
lim(x→0) sin(x)/x = 1 | Special trig limit |
lim(x→0) (cos(x)−1)/x = 0 | Special trig limit |
F'(x) = f'(g(x))·g'(x) | Chain rule |
dy/dx = (dy/du)·(du/dx) | Chain rule (Leibniz) |
Implicit differentiation procedure: differentiate both sides with respect to x, attach dy/dx to every y-term via the chain rule, then solve for dy/dx.
Trigonometric derivatives show up whenever periodic phenomena are modelled: oscillating springs, alternating current, sound waves. The chain rule is essential for any model where one quantity depends on another through an intermediate variable. Implicit differentiation is the starting point for related rates problems (Section 8), where two or more quantities change with time and are linked by a geometric or physical equation.
"The derivative of sin(x) is −cos(x)." The sign is positive: (d/dx)sin(x) = cos(x). It is cos whose derivative picks up the minus sign.
"I can skip the chain rule when the inside is simple." If the inside is anything other than plain x, you need the chain rule. Even sin(2x) requires it: the derivative is 2cos(2x), not cos(2x).
"In implicit differentiation, every term gets dy/dx." Only terms that contain y produce a dy/dx factor. Pure-x terms differentiate normally.
"I need to solve for y before differentiating." That is the whole point of implicit differentiation: you do not solve for y first.
⚠️ You are expected to know all six trig derivatives from memory. The "co" functions always carry a negative.
⚠️ The chain rule appears in nearly every derivative problem from this point on. Forgetting to multiply by the derivative of the inner function is the single most common chain-rule error.
⚠️ Special trig limits (sin(x)/x → 1 and (cos(x)−1)/x → 0) are tested as standalone limit problems. Be comfortable manipulating expressions into the standard form so the limit applies.
⚠️ Implicit differentiation problems typically ask you to find dy/dx and then the equation of the tangent line at a given point. Do not forget to substitute the point's coordinates into dy/dx after solving.
⚠️ The 83rd derivative of sin(x) is a classic pattern-recognition problem: the cycle repeats every 4 derivatives (sin → cos → −sin → −cos), so divide 83 by 4, note the remainder of 3, and the answer is −cos(x).
Fill in the blank: (d/dx) cos(x) = ___.
True or false: (d/dx) tan(x) = sec(x)·tan(x).
Fill in the blank: If h(x) = (3x + 1)^4, then h'(x) = ___.
True or false: When differentiating x² + y² = 25 implicitly, the x² term produces 2x·(dy/dx).
What is lim(x→0) sin(7x)/(7x)?
Answers: 1. −sin(x). 2. False (it is sec²(x)). 3. 4(3x + 1)³ · 3 = 12(3x + 1)³. 4. False (2x has no dy/dx; only y² produces a dy/dx factor). 5. 1.
Q: Differentiate f(x) = sin(x)·cos(x).
A: Product rule: f'(x) = cos(x)·cos(x) + sin(x)·(−sin(x)) = cos²(x) − sin²(x) = cos(2x).
Q: Find the derivative of g(x) = (x² + 1)^(3/2).
A: Chain rule: g'(x) = (3/2)(x² + 1)^(1/2) · 2x = 3x·√(x² + 1).
Q: Find lim(x→0) sin(5x)/(3x).
A: Rewrite as (5/3)·[sin(5x)/(5x)]. As x → 0, sin(5x)/(5x) → 1, so the limit is 5/3.
Q: Use implicit differentiation to find dy/dx for x² + 4xy + y² = 13.
A: Differentiate: 2x + 4y + 4x·(dy/dx) + 2y·(dy/dx) = 0. Solve: dy/dx = −(2x + 4y)/(4x + 2y) = −(x + 2y)/(2x + y).
Q: Find the tangent line to the curve x³ + y³ = 6xy at the point (3, 3).
A: From the implicit differentiation worked above, dy/dx at (3, 3) = −1. Tangent line: y − 3 = −1(x − 3), so y = −x + 6.
The chain rule is essential for the related rates problems in Section 8, where you differentiate an equation relating two or more changing quantities with respect to time. Implicit differentiation is simply the chain rule applied to y as a function of x, so these two topics reinforce each other. The trig derivatives will reappear in integration (Chapter 4) when you learn antiderivatives.
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