AP Calculus Midterms

lim x→0 sin(kx)/x

k

lim x→0 tan(kx)/x

k

lim x→0 (1 − cos x)/x

0

tan²x + 1

sec²x

1 + cot²x

csc²x

sin 2x

2 sin x cos x

cos 2x

cos²x − sin²x

cos 2x in terms of sin only

1 − 2sin²x

cos 2x in terms of cos only

2cos²x − 1

Squeeze bounds that give different numbers

Cannot be determined

Continuity at a: three checks

f(a) exists; the limit exists; they are equal

Rational function: cancelled factor and leftover denominator factor

Hole and vertical asymptote

Rational functions can never have a

Jump discontinuity

Constant that makes a piecewise f continuous

Left piece = right piece at the join (answer the constant and not the height)

Horizontal asymptote of a rational function

Top < bottom gives 0; equal gives leading-coefficient ratio; top > bottom gives none

Which wins at +∞ between eˣ and a polynomial

eˣ (and a bigger base beats a smaller one)

Domain says x > 0 for a horizontal asymptote

Only +∞ counts

IVT requires

Continuous on the closed interval (defined or increasing is not enough)

Fewest zeros from a table

Count the sign changes

Best estimate of an instantaneous rate

Secant over the narrowest symmetric interval around the point

Limit definition of f′(a)

lim h→0 (f(a+h) − f(a))/h

Limit that looks like a derivative

Find a and f; then check the subtracted number equals f(a)

Not differentiable at

Corner; cusp; break; vertical tangent (a horizontal tangent IS differentiable)

Piecewise differentiable at the join

Check continuity first; then match the slopes

d/dx tan x

sec²x

d/dx sec x

sec x tan x

d/dx cot x

−csc²x

d/dx csc x

−csc x cot x

Trig derivatives starting with co

Get a minus

Quotient rule

(v·u′ − u·v′)/v²

Simplify before the quotient rule only when

The denominator is a single term

sin at π/6 and π/4 and π/3

1/2 and √2/2 and √3/2

cos at π/6 and π/4 and π/3

√3/2 and √2/2 and 1/2

tan at π/6 and π/4 and π/3

√3/3 and 1 and √3

Positive trig functions by quadrant

I all; II sin; III tan; IV cos

Chain rule from a table

h′(a) = f′(g(a))·g′(a); find g(a) first

Horizontal and vertical tangents of an implicit curve

Numerator of y′ = 0 and denominator of y′ = 0

Second derivative of an implicit curve

Differentiate y′; substitute y′; simplify with the original equation

Inverse function derivative

(f⁻¹)′(b) = 1/f′(a) where f(a) = b

d/dx arcsin x

1/√(1 − x²)

d/dx arccos x

−1/√(1 − x²)

d/dx arctan x

1/(1 + x²)

d/dx arccot x

−1/(1 + x²)

d/dx arcsec x

1/(|x|√(x² − 1))

d/dx arccsc x

−1/(|x|√(x² − 1))

Ranges of arcsin and arccos and arctan

[−π/2 to π/2] and [0 to π] and (−π/2 to π/2)