TTK ALL CALC AB/BC

sin² θ + cos² θ=

1

1 + tan² θ =

• sec² θ

1 + cot² θ

csc² θ

sin 2θ =

2sin θ ⋅ cos θ

cos² θ – sin² θ

cos2θ

2cos² θ – 1

cos 2θ

1 – 2 sin² θ

cos2θ

sin²x =

(1-cos(2x)) / 2

cos²x =

(1+cos(2x)) / 2

arcsin output?

[-pi/2 , pi/2]

arccos ouput?

[0,pi]

arctan output?

(-pi/2 , pi/2)

arccot output?"

(0,pi)

arcsec output?

[0,pi/2) U {pi/2, pi]

arccsc output?

[-pi/2, 0 ) U (0, pi/2]

limₓ→c b =

b

lim x→c [b · f(x)] =

b · lim x→c f(x) = b · L

lim x→c [f(x) ± g(x)] =

lim x→c f(x) ± lim x→c g(x) = L ± K

lim x→c [f(x) · g(x)] =

lim x→c f(x) · lim x→c g(x) = L · K

lim x→c [f(x) / g(x)] =`

(lim x→c f(x)) / (lim x→c g(x)) = L / K If K ≠ 0

lim x→c [f(x)]ⁿ =

[lim x→c f(x)]ⁿ = Lⁿ

lim x→0 [sin(ax) / ax] =

1

lim x→0 [ax / sin(ax)] =

1

lim x→0 [(1 − cos x) / x] =

0

lim x→0 [cos x / x] =

0

if lim x→c⁻ f(x) = ±∞ or lim x→c⁺ f(x) = ±∞,

then x = c is a vertical asymptote.

If lim x→±∞ f(x) = d,

then y = d is a horizontal asymptote.

constant / approaching 0 =

±∞ (be careful of sign!)

constant / approaching ±∞ =

0

N < D

lim x→∞ f(x) =
lim x→−∞ f(x) =

0

N = D

lim x→∞ f(x) =
lim x→−∞ f(x) =

ratio of leading coefficients

N > D

lim x→∞ f(x) =
lim x→−∞ f(x) =

±∞ (be careful of sign!)

No horizontal asymptote

lim x→c f(x) = f(c) = L if and only if…

f(x) is continuous at x = c.

Removable dis:

lim x→c f(x) ≠ f(c

Jump (Non-Removable) dis

lim x→c⁻ f(x) = L ≠ M = lim x→c⁺ f(x)

Infinite (Non-Removable) dis:

lim x→c⁻ f(x) = ±∞ OR lim x→c⁺ f(x) = ±∞

Oscillating (Non-Removable)

• Function value oscillates near x = c.

If f(x) is continuous on [a, b],

then f(x) takes on every value between f(a) and f(b).

limit derivative for all x

lim h→0 [f(x + h) − f(x)] / h

limit derivative for specific x (x=a)

lim x→a [f(x) − f(a)] / (x − a)

limit derivative for specific x (x=a) with h>0

lim h→0 [f(a + h) − f(a)] / h

A function f(x) is differentiable at x = c, if the derivative from the left of x = c is equal to the derivative from the right of x = c OR

lim x→c⁻ [f(x) − f(c)] / (x − c) = lim x→c⁺ [f(x) − f(c)] / (x − c) OR lim x→c⁻ f′(x) = lim x→c⁺ f′(x)

Horizontal tangents

occur where the numerator of the derivative equals zero (but the function value is still defined).

Vertical tangents

occur where the denominator of the derivative equals zero.

Let f be defined at c. If f′(c) = 0 or f′(c) is undefined,

then c is called a critical value of f.

At a critical point c:

1. If f′ changes sign from positive to negative at c,

then f has a relative maximum value at c.

If f′ changes sign from negative to positive at c,

then f has a relative minimum value at c.

If f′ does not change sign at c,

then f has no relative extreme value at c.

on interval [a,b]
• If f′ < 0 for x > a, then f has a __________ at x = a. In other words, if the function decreases from the left endpoint, then

the left endpoint is a relative maximum.

on interval [a,b] If f′ > 0 for x > a, then f has a ______ at x = a. In other words, if the function increases from the left endpoint, then

then the left endpoint is a relative minimum.

If f′ < 0 for x < b, then f has a ________ at x = b. In other words, if the function decreases into the right endpoint, then

the right endpoint is a relative minimum

f f′ > 0 for x < b, then f has a _______ at x = b. In other words, if the function increases into the right endpoint, then

the right endpoint is a relative maximum.

A function f is increasing when

f ′ > 0 (positive).

A function f is decreasing when

f ′ < 0 (negative).

If f ′′ > 0 on (a, b), then

f ′ is increasing and f is concave up on (a, b).

If f ′′ < 0 on (a, b), then

f ′ is decreasing and f is concave down on (a, b).

f has a point of inflection at x if

f ′′ changes sign at x.

If f ′(c) = 0 and f ′′(c) > 0, then

f(c) is a relative minimum of f.

If f ′(c) = 0 and f ′′(c) < 0, then

f(c) is a relative maximum of f.

If f ′(c) = 0 and f ′′(c) = 0, then no conclusion regarding relative extrema is

possible.

You must use the 1st Derivative sign chart instead.

d/dx [f(g(x))] =

f′(g(x)) · g′(x)

dy/dx =

(dy/du) · (du/dx)

d/dx [sin u] =

u′ · cos u

d/dx [cos u] =

u′ · (−sin u)

d/dx [tan u] =

u′ · sec² u

d/dx [cot u] =

u′ · (−csc² u)

d/dx [sec u] =

u′ · sec u tan u

d/dx [csc u] =

u′ · (−csc u cot u)

d/dx [uⁿ] =

u′ · nuⁿ⁻¹

d/dx [constant] =

0

d/dx [a · f(x)] =

a · d/dx [f(x)]

Velocity is the derivative of ?: v(t) = ?′(t)

postion, s (variable depends on context)

When velocity > 0, object is moving in a

positive direction (right or up for linear

motion)

When velocity < 0, object is moving in

a negative direction (left or down for linear

motion)

When velocity equals 0,

the object is at rest

Acceleration is the derivative of ? and the 2nd derivative of ?:

a(t) = v′(t) = s′′(t)

velocity, postion

When acceleration > 0, the

velocity of the object is increasing. (NOT SPEED)

When acceleration < 0,

the velocity of the object is decreasing. (NOT SPEED)

Speed =

|velocity|

Speeding up when

velocity and acceleration have the same sign.

Slowing down when

velocity and acceleration have opposite signs.

d/dx(e^u) =

u' · e^u

d/dx(a^u) =

u' · a^u · ln(a)

d/dx(ln u) =

u' / u

d/dx(log_a u) =

u' / (u · ln(a))

If f(x) is continuous on [a, b] and differentiable on (a, b), then

Mean Value Theorem (MVT): there exists a value c in (a, b) such that: f'(c) = [f(b) - f(a)] / (b - a)

If f(x) is continuous on [a, b], differentiable on (a, b), AND f(a) = f(b), then

Rolle's Theorem: there exists a value c in (a, b) such that: f'(c) = [f(b) - f(a)] / (b - a) = 0

If lim(x→a) [f(x)/g(x)] is indeterminate , then:

lim(x→a) [f(x)/g(x)] = lim(x→a) [f'(x)/g'(x)]

d/dx[f(g(x))] =

f'(g(x)) · g'(x) V1 chain rule

dy/dx =

(dy/du) · (du/dx) V2 Chain rule (less common)

d/dx[f(x)g(x)] =

g(x)f'(x) + f(x)g'(x)

d/dx[f(x)/g(x)] = ? OR ?

[g(x)f'(x) - f(x)g'(x)] / [g(x)]² OR = (Low · dHigh - High · dLow) / (Low · Low)

If lim x→a f(x)/g(x) is indeterminate, then

lim x→a f(x)/g(x) = lim x→a f'(x)/g'(x)

Steps for Implicit Differentiation:

1. Differentiate both sides with respect to x. Don’t forget the chain rule! When taking the derivative of a term with a y, the chain rule will create a dy/dx.

2. Collect all terms containing dy/dx on one side of the equation. Get all other terms to the other side of the equation.

3. Factor out dy/dx.

4. Solve for dy/dx using division.

Formula for Evaluating the Derivative of an Inverse Function

(f⁻¹)'(x) = 1 / f'(f⁻¹(x))

d/dx [arcsin u] =

u' / √(1 − u²)

d/dx [arctan u] =

u' / (1 + u²)

d/dx [arcsec u] =

u' / (|u|√(u² − 1))

d/dx [arccos u] =

−u' / √(1 − u²)

d/dx [arccot u] =

−u' / (1 + u²)