Source: Unit 3 Calc AB Review, Texas A&M University
Tags: chain rule, composite functions, inverse trig derivatives, arctan, arcsin, arccos, arcsec, differentiation, AP Calculus AB
Unit 3 brings together the chain rule for differentiating composite functions with the specific derivative formulas for inverse trigonometric functions. Nearly every problem on this unit layers one technique on top of the other, so fluency with both is essential. If you can spot the "outer vs. inner" structure of a composition and recall the inverse trig derivative templates, the rest is mechanical.
Chain rule
If y = f(g(x)), then dy/dx = f'(g(x)) · g'(x). In words: differentiate the outer function, leave the inner function alone, then multiply by the derivative of the inner function.
Composite function
A function built by feeding one function's output into another, written f(g(x)). The chain rule exists precisely because of these.
Inner function / outer function
In f(g(x)), g(x) is the inner function and f is the outer function. Identifying which is which is the entire first step of the chain rule.
Inverse trigonometric functions (inverse trig, arc-functions)
The functions arcsin, arccos, arctan, arcsec, arccsc, arccot. Each "undoes" its corresponding trig function on a restricted domain and has its own derivative formula.
For any composition y = f(g(x)):
Differentiate the outer function as if the inner function were just a variable
Multiply by the derivative of the inner function
This can nest: if you have f(g(h(x))), apply the rule from the outside in, picking up a factor at each layer
Shorthand with substitution: let u = g(x), so y = f(u), then dy/dx = (dy/du) · (du/dx)
Exponential compositions: d/dx [e^(u)] = e^(u) · u'
Trig compositions: d/dx [sin(u)] = cos(u) · u', and similarly for cos, tan, sec, etc.
Power compositions: d/dx [u^n] = n · u^(n−1) · u'
Logarithmic compositions: d/dx [ln(u)] = (1/u) · u'
Nested trig powers: d/dx [sin^n(u)] = n · sin^(n−1)(u) · cos(u) · u' (chain rule applied twice)
These are the core templates. Every problem on the review layers the chain rule on top of one of these.
d/dx [arcsin(u)] = 1 / √(1 − u²) · u'
d/dx [arccos(u)] = −1 / √(1 − u²) · u'
d/dx [arctan(u)] = 1 / (1 + u²) · u'
d/dx [arcsec(u)] = 1 / (|u| · √(u² − 1)) · u'
d/dx [arccsc(u)] = −1 / (|u| · √(u² − 1)) · u'
d/dx [arccot(u)] = −1 / (1 + u²) · u'
Note the pattern: arcsin and arccos share the same denominator √(1 − u²) but differ by sign. Arctan and arccot share (1 + u²). Arcsec and arccsc share |u|√(u² − 1).
Some problems combine an inverse trig function with another factor, e.g. x² · arcsin(x). Here you need the product rule first, then the inverse trig derivative formula inside one of the terms.
Chain rule (general): d/dx [f(g(x))] = f'(g(x)) · g'(x)
The six inverse trig derivatives (with chain rule built in, where u = u(x)):
Function | Derivative |
|---|---|
arcsin(u) | u' / √(1 − u²) |
arccos(u) | −u' / √(1 − u²) |
arctan(u) | u' / (1 + u²) |
arccot(u) | −u' / (1 + u²) |
arcsec(u) | u' / (|u| · √(u² − 1)) |
arccsc(u) | −u' / (|u| · √(u² − 1)) |
⚠️ The chain rule is involved in virtually every derivative problem beyond the basics. If you only learn one technique well, make it this one.
⚠️ Forgetting to multiply by the inner derivative (u') is the single most common chain rule error. If your answer is off by a constant factor, check this first.
⚠️ arcsin and arccos derivatives differ only by sign. Mixing them up is a classic exam pitfall.
⚠️ The arcsec derivative has an absolute value in the denominator, |u|. Leaving this out can cost marks on free response.
⚠️ When a problem combines the product rule with an inverse trig function (like x² · arcsin(x)), students often forget to apply the product rule entirely and jump straight to the inverse trig formula.
Q: Find f'(x) if f(x) = arctan(4x).
A: Using d/dx [arctan(u)] = u'/(1 + u²) with u = 4x, u' = 4: f'(x) = 4 / (1 + 16x²)
Q: Find f'(x) if f(x) = x² · arcsin(x).
A: Product rule: f'(x) = 2x · arcsin(x) + x² · (1/√(1 − x²)) = 2x · arcsin(x) + x² / √(1 − x²)
Q: Find f'(x) if f(x) = arccos(1/x).
A: Let u = 1/x = x^(−1), so u' = −x^(−2) = −1/x². f'(x) = (−1/√(1 − u²)) · u' = (−1/√(1 − 1/x²)) · (−1/x²) = 1 / (x² · √(1 − 1/x²)) Simplify: √(1 − 1/x²) = √((x² − 1)/x²) = √(x² − 1)/|x| So f'(x) = 1 / (|x| · √(x² − 1))
Q: Find f'(x) if f(x) = arctan(x / √(1 − x²)).
A: Let u = x / √(1 − x²). Using the quotient rule on u: u' = [√(1 − x²) · 1 − x · (−x/√(1 − x²))] / (1 − x²) = [√(1 − x²) + x²/√(1 − x²)] / (1 − x²) = [(1 − x² + x²)/√(1 − x²)] / (1 − x²) = 1 / (1 − x²)^(3/2)
Also, 1 + u² = 1 + x²/(1 − x²) = 1/(1 − x²).
So f'(x) = u' / (1 + u²) = [1/(1 − x²)^(3/2)] / [1/(1 − x²)] = 1/√(1 − x²)
(Note: this result makes sense, because arctan(x/√(1 − x²)) = arcsin(x) on (−1, 1).)
Q: Find the derivative of arcsec(3x + x³).
A: Let u = 3x + x³, so u' = 3 + 3x². d/dx [arcsec(u)] = u' / (|u| · √(u² − 1)) = (3 + 3x²) / (|3x + x³| · √((3x + x³)² − 1))
Q: Find d/dx [e^(3x² − 6x + 1)].
A: Chain rule with outer = e^u, inner u = 3x² − 6x + 1, u' = 6x − 6: = (6x − 6) · e^(3x² − 6x + 1)
Q: Find d/dx [sin³(5x² + 2x)].
A: Two layers of chain rule. Outer: u³ with u = sin(5x² + 2x). Inner of inner: 5x² + 2x. = 3 sin²(5x² + 2x) · cos(5x² + 2x) · (10x + 2)
Q: Find d/dx [ln(√(4x² + 1))].
A: Simplify first: ln(√(4x² + 1)) = (1/2) ln(4x² + 1). d/dx = (1/2) · 8x / (4x² + 1) = 4x / (4x² + 1)
Q: Find d/dx [e^(3x²) + 5x].
A: d/dx [e^(3x²)] = 6x · e^(3x²), and d/dx [5x] = 5. = 6x · e^(3x²) + 5
Q: Find d/dx [sec(4x³ − x)].
A: d/dx [sec(u)] = sec(u)tan(u) · u' with u = 4x³ − x, u' = 12x² − 1: = (12x² − 1) · sec(4x³ − x) · tan(4x³ − x)
Q: Find d/dx [tan(√(5x² − 1))].
A: Outer: tan(u), u = √(5x² − 1) = (5x² − 1)^(1/2). u' = (1/2)(5x² − 1)^(−1/2) · 10x = 5x / √(5x² − 1) d/dx = sec²(√(5x² − 1)) · 5x / √(5x² − 1)
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