What You Need to Know The AP Calculus AB/BC formula sheet (the one College Board provides) is a minimal safety net , not a full list of everything you need. Your job is to (1) know exactly what’s on it, (2) know how to combine it with rules you must memorize (especially chain rule , power rule , u-sub , etc.), and (3) avoid sign/interval traps.
Two big ideas:
The sheet gives you core derivative/integral pairs , FTC statements , integration by parts , and (BC) standard Maclaurin series . You still must supply the structure : algebra, trig identities, chain rule, and correct limits/intervals. Critical reminder: The sheet’s formulas are usually for x x x u ( x ) u(x) u ( x ) chain rule (for derivatives) and u-sub (for integrals) are the hidden steps.
Step-by-Step Breakdown Identify the task type Derivative? Integral? Accumulation/area? Series approximation? Match the pattern to a formula on the sheet Example: see sec 2 ( x ) \sec^2(x) sec 2 ( x ) tan ( x ) \tan(x) tan ( x ) See ∫ a x f ( t ) d t \int_a^x f(t)\,dt ∫ a x f ( t ) d t FTC + chain rule . Check for an “inside function” u ( x ) u(x) u ( x ) If differentiating: apply chain rule to the sheet formula. If integrating: look for a derivative factor and do u-sub. Do a quick sign + constant check Indefinite integrals need + C +C + C Trig pairs have common sign errors (especially ∫ sin ( x ) d x \int \sin(x)\,dx ∫ sin ( x ) d x If it’s a series question (BC) Start from a known Maclaurin series on the sheet. Substitute/scale to match the target. State interval of convergence after substitution. B. Micro-worked examples (method in action) FTC + chain rule
Given g ( x ) = ∫ 2 x 3 1 + t 2 d t g(x)=\int_2^{x^3} \sqrt{1+t^2}\,dt g ( x ) = ∫ 2 x 3 1 + t 2 d t Use FTC: d d x ∫ 2 u ( x ) f ( t ) d t = f ( u ( x ) ) u ′ ( x ) \frac{d}{dx}\int_2^{u(x)} f(t)\,dt = f(u(x))\,u'(x) d x d ∫ 2 u ( x ) f ( t ) d t = f ( u ( x )) u ′ ( x ) So g ′ ( x ) = 1 + ( x 3 ) 2 ⋅ 3 x 2 = 3 x 2 1 + x 6 g'(x)=\sqrt{1+(x^3)^2}\cdot 3x^2 = 3x^2\sqrt{1+x^6} g ′ ( x ) = 1 + ( x 3 ) 2 ⋅ 3 x 2 = 3 x 2 1 + x 6 Integration by parts (on the sheet)
Compute ∫ x e x d x \int x e^x\,dx ∫ x e x d x Use ∫ u d v = u v − ∫ v d u \int u\,dv = uv-\int v\,du ∫ u d v = uv − ∫ v d u Pick u = x u=x u = x d v = e x d x dv=e^x dx d v = e x d x d u = d x du=dx d u = d x v = e x v=e^x v = e x Result: x e x − ∫ e x d x = x e x − e x + C = e x ( x − 1 ) + C x e^x-\int e^x\,dx = x e^x - e^x + C = e^x(x-1)+C x e x − ∫ e x d x = x e x − e x + C = e x ( x − 1 ) + C Series substitution (BC)
Want Maclaurin for sin ( 3 x ) \sin(3x) sin ( 3 x ) From sheet: sin ( x ) = ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 ( 2 n + 1 ) ! \sin(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!} sin ( x ) = ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 )! x 2 n + 1 Substitute x ↦ 3 x x\mapsto 3x x ↦ 3 x sin ( 3 x ) = ∑ n = 0 ∞ ( − 1 ) n ( 3 x ) 2 n + 1 ( 2 n + 1 ) ! \sin(3x)=\sum_{n=0}^{\infty}(-1)^n\frac{(3x)^{2n+1}}{(2n+1)!} sin ( 3 x ) = ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 )! ( 3 x ) 2 n + 1 A. Limits & derivative definition (know what it means ) Formula When to use Notes f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} f ′ ( x ) = lim h → 0 h f ( x + h ) − f ( x ) Definition of derivative Use when asked “from first principles,” or to justify differentiability.
B. Core derivative rules (NOT fully given on the sheet—memorize) Rule Formula Notes Power rule d d x x n = n x n − 1 \frac{d}{dx}x^n = n x^{n-1} d x d x n = n x n − 1 Works for any real n n n Constant multiple d d x [ c f ( x ) ] = c f ′ ( x ) \frac{d}{dx}[c\,f(x)] = c\,f'(x) d x d [ c f ( x )] = c f ′ ( x ) Factor constants out. Sum/difference d d x [ f ( x ) ± g ( x ) ] = f ′ ( x ) ± g ′ ( x ) \frac{d}{dx}[f(x)\pm g(x)] = f'(x)\pm g'(x) d x d [ f ( x ) ± g ( x )] = f ′ ( x ) ± g ′ ( x ) Differentiate term-by-term. Product rule ( f g ) ′ = f ′ g + f g ′ (fg)'=f'g+fg' ( f g ) ′ = f ′ g + f g ′ Don’t multiply out if it’s messy. Quotient rule ( f g ) ′ = f ′ g − f g ′ g 2 \left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2} ( g f ) ′ = g 2 f ′ g − f g ′ Often easier to rewrite as f ⋅ g − 1 f\cdot g^{-1} f ⋅ g − 1 Chain rule d d x f ( u ( x ) ) = f ′ ( u ( x ) ) u ′ ( x ) \frac{d}{dx}f(u(x))=f'(u(x))\,u'(x) d x d f ( u ( x )) = f ′ ( u ( x )) u ′ ( x ) The #1 “hidden” step with the formula sheet.
C. Derivatives of common functions (these are the classic “table” items) Function Derivative Notes sin ( x ) \sin(x) sin ( x ) cos ( x ) \cos(x) cos ( x ) cos ( x ) \cos(x) cos ( x ) − sin ( x ) -\sin(x) − sin ( x ) Sign trap. tan ( x ) \tan(x) tan ( x ) sec 2 ( x ) \sec^2(x) sec 2 ( x ) cot ( x ) \cot(x) cot ( x ) − csc 2 ( x ) -\csc^2(x) − csc 2 ( x ) sec ( x ) \sec(x) sec ( x ) sec ( x ) tan ( x ) \sec(x)\tan(x) sec ( x ) tan ( x ) csc ( x ) \csc(x) csc ( x ) − csc ( x ) cot ( x ) -\csc(x)\cot(x) − csc ( x ) cot ( x ) ln ( x ) \ln(x) ln ( x ) 1 x \frac{1}{x} x 1 Domain x > 0 x>0 x > 0 e x e^x e x e x e^x e x a x a^x a x a x ln ( a ) a^x\ln(a) a x ln ( a ) Requires a > 0 a>0 a > 0 a ≠ 1 a\neq 1 a = 1
D. Derivatives of inverse trig (common MCQ/FRQ targets) Function Derivative Notes arcsin ( x ) \arcsin(x) arcsin ( x ) 1 1 − x 2 \frac{1}{\sqrt{1-x^2}} 1 − x 2 1 Domain typically ∣ x ∣ < 1 |x|<1 ∣ x ∣ < 1 arccos ( x ) \arccos(x) arccos ( x ) − 1 1 − x 2 -\frac{1}{\sqrt{1-x^2}} − 1 − x 2 1 Same denominator, negative sign. arctan ( x ) \arctan(x) arctan ( x ) 1 1 + x 2 \frac{1}{1+x^2} 1 + x 2 1 Defined for all real x x x
E. Core integration facts (FTC + antiderivatives) Fact Formula Notes FTC Part 1 d d x ∫ a x f ( t ) d t = f ( x ) \frac{d}{dx}\int_a^x f(t)\,dt=f(x) d x d ∫ a x f ( t ) d t = f ( x ) If upper limit is u ( x ) u(x) u ( x ) u ′ ( x ) u'(x) u ′ ( x ) FTC Part 2 ∫ a b f ( x ) d x = F ( b ) − F ( a ) \int_a^b f(x)\,dx=F(b)-F(a) ∫ a b f ( x ) d x = F ( b ) − F ( a ) Requires F ′ ( x ) = f ( x ) F'(x)=f(x) F ′ ( x ) = f ( x ) Average value f avg = 1 b − a ∫ a b f ( x ) d x f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx f avg = b − a 1 ∫ a b f ( x ) d x Common FRQ for “average value on [ a , b ] [a,b] [ a , b ]
F. Indefinite integrals of trig “pairs” (watch signs) Integrand Antiderivative Notes sin ( x ) \sin(x) sin ( x ) − cos ( x ) + C -\cos(x)+C − cos ( x ) + C Sign trap. cos ( x ) \cos(x) cos ( x ) sin ( x ) + C \sin(x)+C sin ( x ) + C sec 2 ( x ) \sec^2(x) sec 2 ( x ) tan ( x ) + C \tan(x)+C tan ( x ) + C csc 2 ( x ) \csc^2(x) csc 2 ( x ) − cot ( x ) + C -\cot(x)+C − cot ( x ) + C Negative. sec ( x ) tan ( x ) \sec(x)\tan(x) sec ( x ) tan ( x ) sec ( x ) + C \sec(x)+C sec ( x ) + C csc ( x ) cot ( x ) \csc(x)\cot(x) csc ( x ) cot ( x ) − csc ( x ) + C -\csc(x)+C − csc ( x ) + C Negative.
G. Integration techniques you must know (some are on the sheet, some aren’t) Technique Formula / trigger Notes u-sub If you see f ′ ( x ) g ( f ( x ) ) f'(x)\,g(f(x)) f ′ ( x ) g ( f ( x )) Reverse chain rule. Integration by parts ∫ u d v = u v − ∫ v d u \int u\,dv = uv-\int v\,du ∫ u d v = uv − ∫ v d u Choose u u u Basic exponential/log integrals ∫ 1 x d x = ln ∣ x ∣ + C \int \frac{1}{x}\,dx=\ln|x|+C ∫ x 1 d x = ln ∣ x ∣ + C ∫ e x d x = e x + C \int e^x dx=e^x+C ∫ e x d x = e x + C Absolute value for ln ∣ x ∣ \ln|x| ln ∣ x ∣
H. BC-only: standard Maclaurin series (know them cold) Function Maclaurin series Interval of convergence 1 1 − x \frac{1}{1-x} 1 − x 1 ∑ n = 0 ∞ x n = 1 + x + x 2 + ⋯ \sum_{n=0}^{\infty} x^n = 1+x+x^2+\cdots ∑ n = 0 ∞ x n = 1 + x + x 2 + ⋯ ∣ x ∣ < 1 |x|<1 ∣ x ∣ < 1 e x e^x e x ∑ n = 0 ∞ x n n ! \sum_{n=0}^{\infty}\frac{x^n}{n!} ∑ n = 0 ∞ n ! x n All real x x x sin ( x ) \sin(x) sin ( x ) ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 ( 2 n + 1 ) ! \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!} ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 )! x 2 n + 1 All real x x x cos ( x ) \cos(x) cos ( x ) ∑ n = 0 ∞ ( − 1 ) n x 2 n ( 2 n ) ! \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!} ∑ n = 0 ∞ ( − 1 ) n ( 2 n )! x 2 n All real x x x ln ( 1 + x ) \ln(1+x) ln ( 1 + x ) ∑ n = 1 ∞ ( − 1 ) n − 1 x n n \sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n} ∑ n = 1 ∞ ( − 1 ) n − 1 n x n − 1 < x ≤ 1 -1<x\le 1 − 1 < x ≤ 1 arctan ( x ) \arctan(x) arctan ( x ) ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 2 n + 1 \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1} ∑ n = 0 ∞ ( − 1 ) n 2 n + 1 x 2 n + 1 ∣ x ∣ ≤ 1 |x|\le 1 ∣ x ∣ ≤ 1
Series substitution rule: If a series converges for ∣ x ∣ < R |x|<R ∣ x ∣ < R x x x k x kx k x ∣ k x ∣ < R |kx|<R ∣ k x ∣ < R ∣ x ∣ < R ∣ k ∣ |x|<\frac{R}{|k|} ∣ x ∣ < ∣ k ∣ R
Examples & Applications 1. Recognize a “reverse chain rule” integral Compute ∫ 2 x 1 + x 2 d x \int \frac{2x}{1+x^2}\,dx ∫ 1 + x 2 2 x d x
Spot inside: 1 + x 2 1+x^2 1 + x 2 2 x 2x 2 x Let u = 1 + x 2 u=1+x^2 u = 1 + x 2 d u = 2 x d x du=2x\,dx d u = 2 x d x Integral becomes ∫ 1 u d u = ln ∣ u ∣ + C = ln ( 1 + x 2 ) + C \int \frac{1}{u}\,du = \ln|u|+C = \ln(1+x^2)+C ∫ u 1 d u = ln ∣ u ∣ + C = ln ( 1 + x 2 ) + C 2. Use the trig integral pairs correctly Compute ∫ csc ( x ) cot ( x ) d x \int \csc(x)\cot(x)\,dx ∫ csc ( x ) cot ( x ) d x
From sheet: ∫ csc ( x ) cot ( x ) d x = − csc ( x ) + C \int \csc(x)\cot(x)\,dx = -\csc(x)+C ∫ csc ( x ) cot ( x ) d x = − csc ( x ) + C Fast check: derivative of csc ( x ) \csc(x) csc ( x ) − csc ( x ) cot ( x ) -\csc(x)\cot(x) − csc ( x ) cot ( x ) 3. FTC Part 1 with a nontrivial upper limit Let h ( x ) = ∫ − 1 cos ( x ) 1 1 + t 2 d t h(x)=\int_{-1}^{\cos(x)} \frac{1}{1+t^2}\,dt h ( x ) = ∫ − 1 c o s ( x ) 1 + t 2 1 d t h ′ ( x ) h'(x) h ′ ( x )
Use FTC + chain: h ′ ( x ) = 1 1 + ( cos ( x ) ) 2 ⋅ d d x [ cos ( x ) ] h'(x)=\frac{1}{1+(\cos(x))^2}\cdot \frac{d}{dx}[\cos(x)] h ′ ( x ) = 1 + ( c o s ( x ) ) 2 1 ⋅ d x d [ cos ( x )] h ′ ( x ) = 1 1 + cos 2 ( x ) ⋅ ( − sin ( x ) ) = − sin ( x ) 1 + cos 2 ( x ) h'(x)=\frac{1}{1+\cos^2(x)}\cdot (-\sin(x))= -\frac{\sin(x)}{1+\cos^2(x)} h ′ ( x ) = 1 + c o s 2 ( x ) 1 ⋅ ( − sin ( x )) = − 1 + c o s 2 ( x ) s i n ( x ) 4. Build a new series from a known one (BC) Find the first three nonzero terms of the Maclaurin series for ln ( 1 − 2 x ) \ln(1-2x) ln ( 1 − 2 x )
Start: ln ( 1 + x ) = ∑ n = 1 ∞ ( − 1 ) n − 1 x n n \ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n} ln ( 1 + x ) = ∑ n = 1 ∞ ( − 1 ) n − 1 n x n Substitute x ↦ − 2 x x\mapsto -2x x ↦ − 2 x ln ( 1 − 2 x ) = ∑ n = 1 ∞ ( − 1 ) n − 1 ( − 2 x ) n n \ln(1-2x)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{(-2x)^n}{n} ln ( 1 − 2 x ) = ∑ n = 1 ∞ ( − 1 ) n − 1 n ( − 2 x ) n First terms:n = 1 : ( − 1 ) 0 − 2 x 1 = − 2 x n=1: (-1)^0\frac{-2x}{1}=-2x n = 1 : ( − 1 ) 0 1 − 2 x = − 2 x n = 2 : ( − 1 ) 1 4 x 2 2 = − 2 x 2 n=2: (-1)^1\frac{4x^2}{2}=-2x^2 n = 2 : ( − 1 ) 1 2 4 x 2 = − 2 x 2 n = 3 : ( − 1 ) 2 − 8 x 3 3 = − 8 3 x 3 n=3: (-1)^2\frac{-8x^3}{3}=-\frac{8}{3}x^3 n = 3 : ( − 1 ) 2 3 − 8 x 3 = − 3 8 x 3 So ln ( 1 − 2 x ) = − 2 x − 2 x 2 − 8 3 x 3 + ⋯ \ln(1-2x)= -2x-2x^2-\frac{8}{3}x^3+\cdots ln ( 1 − 2 x ) = − 2 x − 2 x 2 − 3 8 x 3 + ⋯ Interval: need − 1 < − 2 x ≤ 1 -1< -2x \le 1 − 1 < − 2 x ≤ 1 − 1 2 ≤ x < 1 2 -\frac{1}{2}\le x < \frac{1}{2} − 2 1 ≤ x < 2 1 Common Mistakes & Traps Forgetting chain rule with sheet derivatives
Wrong: differentiating sin ( 5 x ) \sin(5x) sin ( 5 x ) cos ( 5 x ) \cos(5x) cos ( 5 x ) Right: d d x sin ( 5 x ) = cos ( 5 x ) ⋅ 5 \frac{d}{dx}\sin(5x)=\cos(5x)\cdot 5 d x d sin ( 5 x ) = cos ( 5 x ) ⋅ 5 Fix: circle the “inside” u ( x ) u(x) u ( x ) u ′ ( x ) u'(x) u ′ ( x ) Messing up the signs on trig antiderivatives
Common wrong ones: ∫ sin ( x ) d x = cos ( x ) + C \int \sin(x)dx=\cos(x)+C ∫ sin ( x ) d x = cos ( x ) + C ∫ csc 2 ( x ) d x = cot ( x ) + C \int \csc^2(x)dx=\cot(x)+C ∫ csc 2 ( x ) d x = cot ( x ) + C Fix: memorize the “negative” ones: sin → − cos \sin\to -\cos sin → − cos csc 2 → − cot \csc^2\to -\cot csc 2 → − cot csc cot → − csc \csc\cot\to -\csc csc cot → − csc Dropping + C +C + C
On FRQs, missing + C +C + C Fix: if there are no bounds , there must be + C +C + C Confusing ln ( x ) \ln(x) ln ( x ) ln ∣ x ∣ \ln|x| ln ∣ x ∣
Correct: ∫ 1 x d x = ln ∣ x ∣ + C \int \frac{1}{x}\,dx=\ln|x|+C ∫ x 1 d x = ln ∣ x ∣ + C Fix: absolute value is required because d d x ln ∣ x ∣ = 1 x \frac{d}{dx}\ln|x|=\frac{1}{x} d x d ln ∣ x ∣ = x 1 x ≠ 0 x\ne 0 x = 0 Inverse trig derivative mix-ups
Classic trap: swapping denominators for arcsin ( x ) \arcsin(x) arcsin ( x ) arctan ( x ) \arctan(x) arctan ( x ) Fix: arcsin , arccos \arcsin,\arccos arcsin , arccos 1 − x 2 \sqrt{1-x^2} 1 − x 2 arctan \arctan arctan 1 + x 2 1+x^2 1 + x 2 Using a Maclaurin series but not updating the interval after substitution
Example: ln ( 1 + x ) \ln(1+x) ln ( 1 + x ) − 1 < x ≤ 1 -1<x\le 1 − 1 < x ≤ 1 x ↦ − 2 x x\mapsto -2x x ↦ − 2 x Fix: always re-solve the inequality for the new variable. Treating the geometric series as valid at endpoints
For ∑ x n \sum x^n ∑ x n ∣ x ∣ < 1 |x|<1 ∣ x ∣ < 1 Fix: endpoints are not “maybe”; test them. Misreading FTC Part 1 when the variable is in the lower limit
If G ( x ) = ∫ x a f ( t ) d t G(x)=\int_x^a f(t)\,dt G ( x ) = ∫ x a f ( t ) d t G ′ ( x ) = − f ( x ) G'(x)=-f(x) G ′ ( x ) = − f ( x ) Fix: flip limits to match ∫ a x \int_a^x ∫ a x Memory Aids & Quick Tricks Trick / mnemonic Helps you remember When to use “Sine → Cosine, Cosine → Negative Sine ” d d x sin ( x ) = cos ( x ) \frac{d}{dx}\sin(x)=\cos(x) d x d sin ( x ) = cos ( x ) d d x cos ( x ) = − sin ( x ) \frac{d}{dx}\cos(x)=-\sin(x) d x d cos ( x ) = − sin ( x ) Fast derivative recall “A-S-A vs A-T-A ” arcsin / arccos \arcsin/\arccos arcsin / arccos 1 − x 2 \sqrt{1-x^2} 1 − x 2 arctan \arctan arctan 1 + x 2 1+x^2 1 + x 2 Inverse trig derivatives “LIATE ” (or ILATE) Choose u u u Integration by parts selection “Geometric = ratio inside absolute less than 1 ” For ∑ r n \sum r^n ∑ r n ∣ r ∣ < 1 |r|<1 ∣ r ∣ < 1 Series interval of convergence “Upper limit = plug; chain rule multiplier ” d d x ∫ a u ( x ) f ( t ) d t = f ( u ( x ) ) u ′ ( x ) \frac{d}{dx}\int_a^{u(x)} f(t)dt = f(u(x))u'(x) d x d ∫ a u ( x ) f ( t ) d t = f ( u ( x )) u ′ ( x ) FTC Part 1 problems
Quick Review Checklist You can instantly locate and use the sheet’s trig derivative/integral pairs (with correct signs). You automatically apply chain rule when the input isn’t just x x x You know FTC Part 1 and 2 and handle variable limits (including the negative when x x x You never forget + C +C + C You have inverse trig derivatives nailed: arcsin / arccos \arcsin/\arccos arcsin / arccos 1 − x 2 \sqrt{1-x^2} 1 − x 2 arctan \arctan arctan 1 + x 2 1+x^2 1 + x 2 For BC series, you can write the 6 standard Maclaurin series and adjust intervals after substitution. You can use integration by parts quickly and choose u u u You’ve got this—use the sheet as a trigger, and let your rules (chain rule, u-sub, FTC) do the real work.