3.4 Differentiating Inverse Trigonometric Functions

If y = arcsin(x), then sin(y) = x with y in [−π/2, π/2]. Differentiate implicitly: cos(y)·dy/dx = 1, so dy/dx = 1/cos(y). Use the identity cos^2(y) = 1 − sin^2(y) and sin(y) = x to get cos(y) = sqrt(1 − x^2) (positive because y is in [−π/2, π/2]). Therefore d/dx[arcsin(x)] = 1 / sqrt(1 − x^2), for −1 < x < 1. If you have a composite, use the chain rule: d/dx[arcsin(u(x))] = u′(x) / sqrt(1 − (u(x))^2). This follows the inverse function/implicit-diff ideas in the CED (FUN-3.E.2). For more practice and a topic guide, see Fiveable’s Topic 3.4 study guide (https://library.fiveable.me/ap-calculus/unit-3/differentiating-inverse-trigonometric-functions/study-guide/pzDRhb3oXKntPRoQgLOz) and try practice problems (https://library.fiveable.me/practice/ap-calculus).

Here are the standard derivatives (use the chain rule if your input is u(x)—multiply by u′(x)). These are the forms AP expects (know principal-value/domain notes): - d/dx[arcsin x] = 1 / sqrt(1 − x^2) for |x| < 1 - d/dx[arccos x] = −1 / sqrt(1 − x^2) for |x| < 1 - d/dx[arctan x] = 1 / (1 + x^2) for all x - d/dx[arccot x] = −1 / (1 + x^2) for all x - d/dx[arcsec x] = 1 / (|x| sqrt(x^2 − 1)) for |x| > 1 - d/dx[arccsc x] = −1 / (|x| sqrt(x^2 − 1)) for |x| > 1 Example with chain rule: d/dx[arcsin(u(x))] = u′(x) / sqrt(1 − (u(x))^2). These follow from the inverse-function/implicit-differentiation approach in Topic 3.4 (CED FUN-3.E). For a focused study guide, see Fiveable’s Topic 3.4 study guide (https://library.fiveable.me/ap-calculus/unit-3/differentiating-inverse-trigonometric-functions/study-guide/pzDRhb3oXKntPRoQgLOz). For more practice, try the AP practice bank (https://library.fiveable.me/practice/ap-calculus).

Use the chain rule any time the inverse trig has a function inside it (not just x). In AP language (FUN-3.E.2) you combine the standard derivative of the inverse trig with the derivative of the inner function. For example, if u = g(x): - d/dx[arcsin(u)] = u′ / sqrt(1 − u^2) - d/dx[arctan(u)] = u′ / (1 + u^2) - d/dx[arccos(u)] = −u′ / sqrt(1 − u^2) So for y = arcsin(3x^2) you get y′ = (6x) / sqrt(1 − 9x^4). You can derive these using implicit differentiation or the inverse function theorem (both are in the CED). On the AP exam you’ll be expected to apply this in algebraic and contextual problems (Unit 3, Topic 3.4). For extra examples and practice, check the Topic 3.4 study guide (https://library.fiveable.me/ap-calculus/unit-3/differentiating-inverse-trigonometric-functions/study-guide/pzDRhb3oXKntPRoQgLOz) and the unit page (https://library.fiveable.me/ap-calculus/unit-3). For lots of practice problems, see https://library.fiveable.me/practice/ap-calculus.

They’re related but different: d/dx[sin x] = cos x (a basic derivative from trig). For arcsin x (the inverse of sin on its principal range), you use the inverse-function rule or implicit differentiation: if y = arcsin x then sin y = x, differentiate to get cos y · y' = 1, so y' = 1/cos y. Replace cos y with √(1 − x^2) (since on the principal branch cos y ≥ 0), giving d/dx[arcsin x] = 1 / √(1 − x^2), for −1 < x < 1. Key differences to remember: domains and ranges matter (arcsin’s derivative exists only for |x|<1), you often need the chain rule when the input isn’t just x (e.g., d/dx[arcsin(u(x))] = u′(x)/√(1−u(x)^2)), and you should cite principal-value/range restrictions from the CED. For more AP-aligned explanation and practice, see the Topic 3.4 study guide (https://library.fiveable.me/ap-calculus/unit-3/differentiating-inverse-trigonometric-functions/study-guide/pzDRhb3oXKntPRoQgLOz) and thousands of practice problems (https://library.fiveable.me/practice/ap-calculus).

Use the derivative formula for inverse trig + the chain rule. 1) Know the basic rule: d/dx[arctan(u)] = u' / (1 + u^2). This comes from implicit differentiation / inverse function theorem (CED FUN-3.E). 2) Let u(x) = 2x. Then u'(x) = 2. 3) Apply the rule: d/dx[arctan(2x)] = u'(x) / (1 + (u(x))^2) = 2 / (1 + (2x)^2). 4) Simplify: 2 / (1 + 4x^2). So the derivative is 2/(1+4x^2). This is valid for all real x (arctan has principal value range (−π/2, π/2)). For more on differentiating inverse trig and practice problems aligned to the AP CED, see the Topic 3.4 study guide (https://library.fiveable.me/ap-calculus/unit-3/differentiating-inverse-trigonometric-functions/study-guide/pzDRhb3oXKntPRoQgLOz) and AP Calc practice problems (https://library.fiveable.me/practice/ap-calculus).

Think of y = arccos(x) as the inverse of cos on its principal range 0 ≤ y ≤ π. If y = arccos(x), then cos(y) = x. Differentiate implicitly with respect to x: -sin(y) * dy/dx = 1 → dy/dx = -1 / sin(y). Now express sin(y) in terms of x using the Pythagorean identity sin^2(y) + cos^2(y) = 1. Since cos(y) = x, sin(y) = √(1 - x^2). On the principal range 0 ≤ y ≤ π, sin(y) ≥ 0, so we take the positive square root. Thus d/dx[arccos(x)] = -1 / √(1 - x^2). So the derivative is negative because arccos decreases: as x increases, the angle y in [0, π] that has that cosine value actually gets smaller. This is exactly the sort of inverse-function + chain-rule reasoning in the CED (FUN-3.E.2). For more practice and a quick study guide, check the Topic 3.4 study guide (https://library.fiveable.me/ap-calculus/unit-3/differentiating-inverse-trigonometric-functions/study-guide/pzDRhb3oXKntPRoQgLOz) and Unit 3 overview (https://library.fiveable.me/ap-calculus/unit-3).

Use the chain rule whenever the inverse trig function’s input isn’t just x but a composite u(x). The CED (FUN-3.E) expects you to combine the derivative formula for the inverse trig with the chain rule or inverse-function/implicit methods. Quick rule: - If you have y = arctan(x) → dy/dx = 1/(1 + x^2) (no extra chain work because u(x)=x). - If you have y = arctan(u(x)) → dy/dx = u′(x) · 1/(1 + [u(x)]^2). Same pattern for others: - d/dx[arcsin(u)] = u′/(√(1 − u^2)) - d/dx[arccos(u)] = −u′/(√(1 − u^2)) - d/dx[arcsec(u)] = u′/(|u|√(u^2 − 1)), etc. If you’re unsure, try implicit differentiation: set x = sin(y) for y = arcsin(x) and differentiate both sides, then multiply by y′. That gives the same result and is allowed on the AP exam (Topic 3.4). For more worked examples and practice, see the Topic 3.4 study guide (https://library.fiveable.me/ap-calculus/unit-3/differentiating-inverse-trigonometric-functions/study-guide/pzDRhb3oXKntPRoQgLOz) and thousands of practice problems (https://library.fiveable.me/practice/ap-calculus).

d/dx[arcsec x] = 1 / (|x| sqrt(x^2 − 1)), for |x| > 1. Quick derivation (implicit): let y = arcsec x so sec y = x. Differentiate: sec y tan y · y' = 1 ⇒ y' = 1/(sec y · tan y). Replace sec y with x and tan y with ±√(x^2−1). The sign ambiguity is fixed by taking |x| in the denominator, giving 1/(|x|√(x^2−1)) and restricting to |x|>1 (principal branch). How to remember it: - Think “arcsec is like arccos but with a reciprocal.” arccos' = −1/√(1−x^2); arcsec' ends up positive with |x| and √(x^2−1). - Mnemonic: “sec → |x|”—whenever you see sec in the inverse derivative, expect an absolute x in the bottom. - For compositions use the chain rule: d/dx[arcsec(u)] = u'/(|u|√(u^2−1)). This fits AP FUN-3.E (use chain rule or inverse-function/implicit differentiation). For a topic review, see the Topic 3.4 study guide (https://library.fiveable.me/ap-calculus/unit-3/differentiating-inverse-trigonometric-functions/study-guide/pzDRhb3oXKntPRoQgLOz) and try practice problems (https://library.fiveable.me/practice/ap-calculus).

Think of the inverse-derivative formula: (f^{-1})'(x) = 1 / f'(f^{-1}(x)). For trig inverses you can also do implicit differentiation—same idea, often faster. Example (arcsin): let y = arcsin(x) so sin(y) = x. Differentiate: cos(y)·y' = 1, so y' = 1/cos(y). Replace cos(y) using cos(y) = √(1−sin^2(y)) = √(1−x^2). So d/dx[arcsin(x)] = 1/√(1−x^2). Arccos and arctan (use the inverse formula or implicit diff): - d/dx[arccos(x)] = −1/√(1−x^2) - d/dx[arctan(x)] = 1/(1+x^2) If the input is a function g(x), apply the chain rule: d/dx[arcsin(g(x))] = g'(x) / √(1−(g(x))^2). AP tip: the CED allows using either the inverse function formula or implicit differentiation/chain rule (Topic 3.4 FUN-3.E). For extra practice see the Topic 3.4 study guide (https://library.fiveable.me/ap-calculus/unit-3/differentiating-inverse-trigonometric-functions/study-guide/pzDRhb3oXKntPRoQgLOz) and lots of practice problems (https://library.fiveable.me/practice/ap-calculus).

Use the derivative formula for arcsin and the chain rule. If y = arcsin(3x^2), set u(x) = 3x^2. Then dy/dx = u'(x) / sqrt(1 − [u(x)]^2). Compute u'(x) = 6x, so dy/dx = 6x / sqrt(1 − 9x^4). A few notes useful for AP: this follows FUN-3.E.2 (chain rule + derivative of an inverse trig). The derivative is defined only where the denominator is real—i.e., 1 − 9x^4 > 0 so |x| < (1/3)^(1/2) (actually (1/3)^(1/2) = 1/√3 because 9x^4 < 1 ⇒ x^4 < 1/9 ⇒ |x| < 1/√3). On free-response or multiple-choice, show the chain-rule step (u and u') and mention domain if asked. For extra practice, check the Topic 3.4 study guide (https://library.fiveable.me/ap-calculus/unit-3/differentiating-inverse-trigonometric-functions/study-guide/pzDRhb3oXKntPRoQgLOz) and hit the practice problems page (https://library.fiveable.me/practice/ap-calculus).

Use the inverse-trig derivative formulas when your function is explicitly an inverse trig of something: e.g. y = arcsin(u(x)) or y = arctan(3x+1). Then apply the known formula (dy/dx = u'(x)/√(1−u(x)^2) for arcsin, dy/dx = u'(x)/(1+u(x)^2) for arctan, etc.) and the chain rule (FUN-3.E.2). That’s usually fastest and directly tested on the exam. Use implicit differentiation when y is defined implicitly (an equation mixing x and y) or when solving for y explicitly would be messy or impossible—e.g. x^2 + y^2 = 1 or x = y + sin y. Differentiate both sides with respect to x, then solve for dy/dx. You’ll sometimes combine both: differentiate implicitly, then if you get an inverse trig expression for y, you can use inverse-trig formulas + chain rule. AP note: both approaches come from the inverse function theorem and chain rule and are acceptable on Free-Response and MC (CED FUN-3.E.2). For extra practice, check the Topic 3.4 study guide (https://library.fiveable.me/ap-calculus/unit-3/differentiating-inverse-trigonometric-functions/study-guide/pzDRhb3oXKntPRoQgLOz) and Unit 3 overview (https://library.fiveable.me/ap-calculus/unit-3). For lots of practice problems, see (https://library.fiveable.me/practice/ap-calculus).

Step-by-step process you can use every time (works for arcsin, arctan, etc.): 1. Know the standard derivative forms (or derive them): - d/dx[arcsin x] = 1 / sqrt(1 − x^2) - d/dx[arccos x] = −1 / sqrt(1 − x^2) - d/dx[arctan x] = 1 / (1 + x^2) - d/dx[arcsec x] = 1 / (|x| sqrt(x^2 − 1)) - d/dx[arccsc x] = −1 / (|x| sqrt(x^2 − 1)) - d/dx[arccot x] = −1 / (1 + x^2) 2. If inside is not just x, use the chain rule: for y = arcsin(u(x)), y' = (1 / sqrt(1 − u^2)) * u'(x). Always multiply by u'(x). 3. If you forget a formula, use implicit differentiation: set y = arctrig(u) ⇔ trig(y) = u. Differentiate both sides (use trig derivative) and solve for y'. This also gives the same formulas above. 4. Watch domains/principal values: each inverse trig has a restricted range so the derivative signs make sense on that principal branch (important on the AP for justification). 5. On the AP, show either the direct formula + chain rule or a short implicit differentiation derivation to earn full credit (CED FUN-3.E.2). For a quick review and examples, see the Topic 3.4 study guide (https://library.fiveable.me/ap-calculus/unit-3/differentiating-inverse-trigonometric-functions/study-guide/pzDRhb3oXKntPRoQgLOz). For more practice, use the AP practice set (https://library.fiveable.me/practice/ap-calculus).

Set y = arctan(x). By definition that means tan(y) = x, with y in the principal range (-π/2, π/2). Differentiate both sides implicitly with respect to x: sec^2(y) · dy/dx = 1. So dy/dx = 1 / sec^2(y). Use the trig identity sec^2(y) = 1 + tan^2(y). But tan(y) = x, so sec^2(y) = 1 + x^2. Therefore d/dx[arctan(x)] = dy/dx = 1/(1 + x^2). This uses the inverse-function/implicit-differentiation idea from the CED (FUN-3.E.2). The domain comes from the principal value of arctan (so the derivative formula holds for all real x). For extra practice or AP-style problems, check the Topic 3.4 study guide (https://library.fiveable.me/ap-calculus/unit-3/differentiating-inverse-trigonometric-functions/study-guide/pzDRhb3oXKntPRoQgLOz) and the practice set (https://library.fiveable.me/practice/ap-calculus).

Use the product rule: y = x·arccos(x) so y' = 1·arccos(x) + x·(arccos(x))'. The derivative of arccos(x) (FUN-3.E, chain rule/inverse trig) is -(1)/√(1−x^2) for −1https://library.fiveable.me/ap-calculus/unit-3/differentiating-inverse-trigonometric-functions/study-guide/pzDRhb3oXKntPRoQgLOz) and use the unit review/practice sets (https://library.fiveable.me/ap-calculus/unit-3 and https://library.fiveable.me/practice/ap-calculus).

Keep three things in mind: the algebraic domain (what x can be), the denominator/square-root that can blow up, and the principal-value ranges (they determine sign conventions for arcsec/arccsc). Quick checklist of derivatives and domain rules (then apply chain rule): - d/dx[arcsin x] = 1 / sqrt(1 − x^2), domain: |x| < 1 (derivative → ∞ at x = ±1). Range/principal value: [−π/2, π/2]. - d/dx[arccos x] = −1 / sqrt(1 − x^2), domain: |x| < 1. Range: [0, π]. - d/dx[arctan x] = 1 / (1 + x^2), domain: all real x. Range: (−π/2, π/2). - d/dx[arccot x] = −1 / (1 + x^2), domain: all real x. - d/dx[arcsec x] = 1 / (|x| sqrt(x^2 − 1)), domain: |x| > 1. Range: [0, π], y ≠ π/2. - d/dx[arccsc x] = −1 / (|x| sqrt(x^2 − 1)), domain: |x| > 1. Common errors: forgetting absolute value in arcsec/arccsc, treating endpoints ±1 as allowed for arcsin/arccos derivatives, or missing the negative signs for arccos/arccot/arccsc. Always check whether the inner function makes the denominator zero or inside a sqrt negative, and then multiply by the inner derivative (chain rule). For more AP-aligned review and practice, see the Topic 3.4 study guide (https://library.fiveable.me/ap-calculus/unit-3/differentiating-inverse-trigonometric-functions/study-guide/pzDRhb3oXKntPRoQgLOz) and unit resources (https://library.fiveable.me/ap-calculus/unit-3).