d/dx[tan x] = sec^2 x. Quick derivation: tan x = sin x / cos x. Use the quotient rule: d/dx(tan x) = (cos·cos − sin·(−sin))/cos^2 = (cos^2 + sin^2)/cos^2 = 1/cos^2 = sec^2 x. You can also get it from the identity 1 + tan^2 x = sec^2 x by differentiating both sides. How to memorize it: - Remember the short phrase: “tan → sec squared.” Picture tan turning into sec^2. - Or link identities: tan = sin/cos (quotient rule) → sec^2; and 1 + tan^2 = sec^2 (so derivative of tan involves sec^2). - For compositions, d/dx[tan(u)] = sec^2(u)·u' (chain rule). This matches the AP CED idea of rearranging trig functions and using derivative rules (see Topic 2.10 study guide: https://library.fiveable.me/ap-calculus/unit-2/derivatives-tangent-cotangent-secant-cosecant-functions/study-guide/wkBAhxiFZ1wV1M5mwLwt). For extra practice, try problems at (https://library.fiveable.me/practice/ap-calculus).

Write sec x as (cos x)^(−1) and use the chain rule (not the quotient rule): d/dx[sec x] = d/dx[(cos x)^(−1)] = (−1)(cos x)^(−2) · d/dx[cos x] = (−1)(cos x)^(−2) · (−sin x) = sin x / cos^2 x. Now use trig identities (sin x / cos^2 x = (1/cos x)(sin x / cos x)): sin x / cos^2 x = sec x · tan x. So d/dx[sec x] = sec x tan x. This follows the CED idea of rearranging trig functions (sec x = (cos x)^(−1)) and applying chain/product rules (FUN-3.B.3). For more worked examples and AP-style practice, see the Topic 2.10 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivatives-tangent-cotangent-secant-cosecant-functions/study-guide/wkBAhxiFZ1wV1M5mwLwt) and try problems on Fiveable (https://library.fiveable.me/practice/ap-calculus).

Use trig identities when rewriting the trig function makes differentiation with product/quotient rules easier or gives a neat simplification; use the chain rule when you have a trig function of some inner function. AP CED expects you to be comfortable both rewriting tan/cot/sec/csc (e.g., tan x = sin x / cos x, sec x = 1/cos x) and applying standard derivative rules (FUN-3.B.3). Quick rules of thumb: - If you see sin(g(x)) or cos(g(x)), use the chain rule: d/dx[sin(g(x))] = cos(g(x))·g'(x). - If you see tan x, cot x, sec x, csc x and want the standard forms, either memorize the results (d/dx tan x = sec^2 x, d/dx cot x = −csc^2 x, d/dx sec x = sec x tan x, d/dx csc x = −csc x cot x) or derive them by rewriting (tan = sin/cos) and using quotient/product + trig identities (e.g., 1 + tan^2 x = sec^2 x) to simplify. - If rewriting produces messy algebra, it’s often faster to apply the known derivative plus chain rule: d/dx[sec(3x)] = sec(3x) tan(3x)·3. On the AP exam you’re allowed to rearrange trig functions to apply product/quotient rules (see Topic 2.10). For extra practice and examples, check the Topic 2.10 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivatives-tangent-cotangent-secant-cosecant-functions/study-guide/wkBAhxiFZ1wV1M5mwLwt) and the practice problem bank (https://library.fiveable.me/practice/ap-calculus).

Short answer: d/dx[tan x] = sec^2 x and d/dx[cot x] = −csc^2 x. The main difference is a sign. Why that happens (two quick ways): 1) Use trig identities (fast AP-friendly): since 1 + tan^2 x = sec^2 x, differentiate tan x (or differentiate tan x and use algebra) to get sec^2 x. For cot, 1 + cot^2 x = csc^2 x and differentiating gives the negative sign: d/dx[cot x] = −csc^2 x. 2) Use quotient rule (good to show on the exam): tan x = sin x / cos x. - Quotient rule: (sin'·cos − sin·cos')/cos^2 = (cos^2 + sin^2)/cos^2 = 1/cos^2 = sec^2 x. cot x = cos x / sin x gives (−sin^2 − cos^2)/sin^2 = −1/sin^2 = −csc^2 x. Remember chain rule if the argument isn’t x: d/dx[tan(u)] = sec^2(u)·u' and d/dx[cot(u)] = −csc^2(u)·u'. For more practice and AP-aligned notes, see the Topic 2.10 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivatives-tangent-cotangent-secant-cosecant-functions/study-guide/wkBAhxiFZ1wV1M5mwLwt) and lots of practice problems (https://library.fiveable.me/practice/ap-calculus).

Think of csc x as 1/sin x or (sin x)^(-1). Using the quotient or chain rule makes the negative sign natural. Quotient-rule route: d/dx[1/sin x] = (0·sin x − 1·cos x)/(sin^2 x) = −cos x / sin^2 x. Chain-rule/power-rule route: d/dx[(sin x)^(−1)] = −1·(sin x)^(−2)·cos x = −cos x / sin^2 x. Then use trig identities: −cos x / sin^2 x = −(1/sin x)(cos x / sin x) = −csc x · cot x. So the negative appears because of the derivative of the inner function (sin x) combined with the power −1 from rewriting csc x as a reciprocal. This ties to the AP CED tools: you can use quotient, chain, or power rules and the identities 1/sin x and cot x = cos/sin to get −csc x cot x (see the Topic 2.10 study guide for worked examples: https://library.fiveable.me/ap-calculus/unit-2/derivatives-tangent-cotangent-secant-cosecant-functions/study-guide/wkBAhxiFZ1wV1M5mwLwt). For extra practice, try related problems at https://library.fiveable.me/practice/ap-calculus.

You use the chain rule plus the basic derivative d/dx[tan x] = sec^2 x. Step-by-step: 1. Recognize tan(3x) is a composition: outer function u ↦ tan(u) with u = 3x. 2. Derivative of outer: d/du[tan u] = sec^2 u. 3. Derivative of inner: d/dx[3x] = 3. 4. Chain rule: d/dx[tan(3x)] = (d/du[tan u] at u=3x) * (d/dx[3x]) = sec^2(3x) * 3. 5. Final answer: d/dx[tan(3x)] = 3 sec^2(3x). You can also get this by rewriting tan as sin/cos and using quotient + chain rules, but using the identity d/dx[tan x] = sec^2 x is quicker (CED FUN-3.B.3). For more on derivatives of trig reciprocals and related practice, see the Topic 2.10 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivatives-tangent-cotangent-secant-cosecant-functions/study-guide/wkBAhxiFZ1wV1M5mwLwt) and tons of practice problems (https://library.fiveable.me/practice/ap-calculus).

You should memorize these six basic derivatives for the AP exam (and apply chain/product/quotient rules as needed): - d/dx[sin x] = cos x - d/dx[cos x] = −sin x - d/dx[tan x] = sec^2 x (since tan x = sin x / cos x and 1 + tan^2 x = sec^2 x) - d/dx[cot x] = −csc^2 x (cot x = cos x / sin x and 1 + cot^2 x = csc^2 x) - d/dx[sec x] = sec x · tan x (can be done via sec x = 1/cos x and quotient/product rules) - d/dx[csc x] = −csc x · cot x (from csc x = 1/sin x) If the argument is a function u(x), use the chain rule: d/dx[sin(u)] = cos(u)·u′(x), etc. The CED expects you to rearrange using identities (e.g., tan = sin/cos) and use quotient/product rules when needed (FUN-3.B.3). For a quick review, see the Topic 2.10 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivatives-tangent-cotangent-secant-cosecant-functions/study-guide/wkBAhxiFZ1wV1M5mwLwt) and more unit resources (https://library.fiveable.me/ap-calculus/unit-2). For extra practice, try problems at https://library.fiveable.me/practice/ap-calculus.

You rewrite sec(x) as 1/cos(x) (or cos(x)^{-1}) so you can apply rules you already know—quotient rule or chain rule—instead of trying to memorize a separate rule. For example, using the chain rule on cos(x)^{-1}: d/dx[cos(x)^{-1}] = −1·cos(x)^{-2}·(−sin x) = sin x / cos^2 x = (1/cos x)(sin x / cos x) = sec x · tan x. So rewriting makes the differentiation a straightforward application of FUN-3.B skills (quotient/product/chain rules) listed in the CED, and it connects the result to trig identities (sin/cos = tan, 1/cos = sec). On the AP exam you’ll often be expected to rearrange trig functions using identities before differentiating (see Topic 2.10 study guide: https://library.fiveable.me/ap-calculus/unit-2/derivatives-tangent-cotangent-secant-cosecant-functions/study-guide/wkBAhxiFZ1wV1M5mwLwt). For extra practice, use the unit review (https://library.fiveable.me/ap-calculus/unit-2) and over 1,000 practice problems (https://library.fiveable.me/practice/ap-calculus).

Yes—use the product rule. x^2 and tan(x) are both differentiable, so for f(x) = x^2·tan(x): f'(x) = (x^2)'·tan(x) + x^2·(tan(x))' = 2x·tan(x) + x^2·sec^2(x). Remember the derivative fact from the CED: d/dx[tan x] = sec^2 x (use tan x = sin x / cos x or the identity 1 + tan^2 x = sec^2 x if you like). If you had tan(kx) instead of tan(x), you’d also apply the chain rule: d/dx[tan(kx)] = k·sec^2(kx). This fits FUN-3.B (products and trig derivatives) on the AP CED. For more practice and worked examples on tangent/secant/cotangent/cosecant derivatives, see the Topic 2.10 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivatives-tangent-cotangent-secant-cosecant-functions/study-guide/wkBAhxiFZ1wV1M5mwLwt) and try extra problems at (https://library.fiveable.me/practice/ap-calculus).

Use whichever gives a cleaner derivative with fewer steps. Guiding rules tied to the CED: - If the function is one of the standard trig ratios (tan, cot, sec, csc), rewrite using identities so you can apply known derivatives (this is what FUN-3.B.3 encourages). Example: tan x = sin x / cos x—you can do quotient rule, but it’s faster to note 1 + tan^2 x = sec^2 x or differentiate tan x by rewriting as sin·cos⁻¹ and using product/chain to obtain sec^2 x. For sec x = 1/cos x it’s usually easiest to write sec x = (cos x)⁻¹ and use chain rule to get sec x tan x. - Use quotient rule when you actually have a messy quotient of two nonstandard functions (or when rewriting makes it more complicated). Quotient rule is always valid but often longer for trig because identities give compact forms and avoid extra algebra. Quick checklist for AP work: if the problem involves a single trig ratio (tan, cot, sec, csc) use the identity/rewrite + chain/product rules; if it’s a complicated quotient of unrelated functions, use quotient rule. For examples and practice aligned to Topic 2.10, see the study guide (https://library.fiveable.me/ap-calculus/unit-2/derivatives-tangent-cotangent-secant-cosecant-functions/study-guide/wkBAhxiFZ1wV1M5mwLwt) and extra practice (https://library.fiveable.me/practice/ap-calculus).

Derivative formulas: - d/dx[tan x] = sec^2 x - d/dx[cot x] = −csc^2 x Why they differ: tan x = sin x / cos x and cot x = cos x / sin x. Using the quotient rule (or the identity 1 + tan^2 x = sec^2 x and 1 + cot^2 x = csc^2 x) gives the results. Quick derivation for cot x: cot x = cos x / sin x d/dx[cot x] = (−sin x · sin x − cos x · cos x) / sin^2 x = −(sin^2 x + cos^2 x)/sin^2 x = −1/sin^2 x = −csc^2 x. Remember the chain rule: d/dx[cot(u(x))] = −csc^2(u(x)) · u′(x) (and similarly for tan: d/dx[tan(u)] = sec^2(u)·u′(x)). This matches the CED identities and derivative rules in Topic 2.10. For a compact review and practice problems on these derivatives, see the Topic 2.10 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivatives-tangent-cotangent-secant-cosecant-functions/study-guide/wkBAhxiFZ1wV1M5mwLwt) and more practice (https://library.fiveable.me/practice/ap-calculus).

Let u = 2x + 1. You can use the known derivative d/dx[csc u] = −csc u · cot u, then apply the chain rule: d/dx[csc(2x + 1)] = (d/d u [csc u]) · (du/dx) = (−csc u · cot u) · (2) = −2 csc(2x + 1) cot(2x + 1). So the derivative is −2 csc(2x+1)cot(2x+1). This uses the identity csc x = 1/sin x and the chain rule (FUN-3.B.3 in the CED). For more worked examples on derivatives of trig reciprocal functions, check the Topic 2.10 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivatives-tangent-cotangent-secant-cosecant-functions/study-guide/wkBAhxiFZ1wV1M5mwLwt) and try practice problems (https://library.fiveable.me/practice/ap-calculus).

Take f(x) = tan x · sec x. Use the product rule: (uv)' = u'v + uv'. 1) u = tan x so u' = sec^2 x (CED: derivative of tan x = sec^2 x). v = sec x so v' = sec x tan x (CED: derivative of sec x = sec x tan x). 2) f'(x) = (sec^2 x)(sec x) + (tan x)(sec x tan x) = sec^3 x + sec x · tan^2 x. 3) Factor sec x: f'(x) = sec x (sec^2 x + tan^2 x). 4) Optional simplification using 1 + tan^2 x = sec^2 x (CED identity): tan^2 x = sec^2 x − 1, so sec^2 x + tan^2 x = 2 sec^2 x − 1, giving f'(x) = sec x (2 sec^2 x − 1). Either sec^3 x + sec x tan^2 x or sec x(2 sec^2 x − 1) is correct. This uses the product rule (FUN-3.B) and trig identities; for another route you could rewrite as sin x / cos^2 x and use the quotient or chain rules. For a quick topic review see the Fiveable study guide (https://library.fiveable.me/ap-calculus/unit-2/derivatives-tangent-cotangent-secant-cosecant-functions/study-guide/wkBAhxiFZ1wV1M5mwLwt) and practice problems (https://library.fiveable.me/practice/ap-calculus).

Most likely your calculator is in degree mode or using a numeric-approximation routine—both will give answers that look “wrong” compared to the standard derivatives you learn in class. On the AP exam and in calculus, derivatives of trig functions (e.g., d/dx[tan x] = sec^2 x, d/dx[cot x] = −csc^2 x, d/dx[sec x] = sec x·tan x, d/dx[csc x] = −csc x·cot x) are defined when x is in radians. If your calculator is set to degrees, the numeric derivative includes a scaling factor (π/180) and won’t match the symbolic/radian result. Other things to check: - Are you using symbolic vs numerical differentiation? Some calculators return approximate slopes, not exact algebraic forms. - Parentheses/argument errors (e.g., tan(x) vs tan x) change results. - If you applied the chain rule, remember to multiply by the inner derivative. For AP prep, always work in radian mode on calculus problems (AP allows calculators only on specific parts—know which ones). Review Topic 2.10 and trig-derivative identities on the Fiveable study guide (https://library.fiveable.me/ap-calculus/unit-2/derivatives-tangent-cotangent-secant-cosecant-functions/study-guide/wkBAhxiFZ1wV1M5mwLwt) and try practice problems (https://library.fiveable.me/practice/ap-calculus) to get comfortable.

Short answer: the odd ones are negative. Derivatives: - d/dx[tan x] = sec^2 x (positive) - d/dx[cot x] = −csc^2 x (negative) - d/dx[sec x] = sec x·tan x (positive) - d/dx[csc x] = −csc x·cot x (negative) Why the sign pattern? Think how each function is built: - tan = sin/cos → differentiate (quotient rule) or use 1 + tan^2 = sec^2 to get a positive sec^2. - cot = cos/sin → quotient rule yields a negative (or use 1 + cot^2 = csc^2 and differentiate). - sec = 1/cos → d(1/cos) = (sin)/(cos^2) = sec·tan (no minus). - csc = 1/sin → d(1/sin) = −cos/(sin^2) = −csc·cot (minus from derivative of sin). Mnemonic: tan/sec → positive; cot/csc → negative. Use quotient or reciprocal + chain/product rules (CED FUN-3.B keywords). For extra practice, see the Topic 2.10 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivatives-tangent-cotangent-secant-cosecant-functions/study-guide/wkBAhxiFZ1wV1M5mwLwt) and thousands of practice problems at (https://library.fiveable.me/practice/ap-calculus).