math 20a final

lim_x→0(sinx)/x

1

lim_x→0(1-cosx)/x

0

d/dx [sinx]

cosx

d/dx [cosx]

-sinx

d/dx [tanx]

sec²x

d/dx [secx]

secxtanx

d/dx [cscx]

-cscxcotx

d/dx [cotx]

-csc²x

d/dx [e^x]

e^x

d/dx [lnx]

1/x

d/dx [a^x]

a^xlna

d/dx [log_ax]

1/(x lna)

range of sin^-1x

[-π/2, π/2]

range of cos^-1x

[0, π]

range of tan^-1x

(-π/2, π/2)

range of cot^-1x

(0, π)

range of sec^-1x

[0, π/2)∩(π/2, π]

range of csc^-1x

[-π/2, 0)∩(0, π/2]

d/dx sin^-1 x

\frac{1}{\sqrt{1-x^2}}

\frac{d}{dx}\cos^{-1}x

-\frac{1}{\sqrt{1-x^2}}

d/dx tan^-1 x

\frac{1}{1+x^2}

d/dx cot^-1 x

-\frac{1}{1+x^2}

d/dx sec^-1 x

\frac{1}{x\sqrt{x^2-1}}

d/dx csc^-1 x

-\frac{1}{x\sqrt{x^2-1}}

d/dx sin^-1 (x/a)

\frac{1}{\sqrt{a^2-x^2}}

d/dx tan^-1 (x/a)

\frac{a}{a^2+x^2}

sin²x + cos²x =

1

sec²x-tan²x =

1

csc²x-cot²x =

1

even trig functions

cosine, secant; f(-x) = f(x)

odd trig functions

sine, tangent, etc; f(-x) = -f(x)

extreme value of f

f(c) is the absolute minimum or absolute maximum of f on interval I

local extreme values of f

f(c) is the local minimum or local maximum of f over some open interval of c

critical point of f

either f’(c)=0 or f’(c) DNE

Rolle’s Theorem

Let f be continuous on [a, b] and differentiable on (a, b). If f(a)=f(b), then there exists c ε (a, b) such that f’(c) = 0.

Mean Value Theorem (MVT)

Let x be continuous on [a, b] and differentiable on (a, b). Then there exists some c ε (a, b) such that f’(c) = f(b)-f(a) / b-a

Cauchy’s MVT

Assume that f(x) and g(x) are continuous on [a, b] and differentiable on (a, b), and g(x) =/= g(b). Then there exists some c ε (a, b) such that the ratio of tangent lines of f, g is equal to the ratio of slopes of the secant lines of f, g.

f is concave up (convex) when…

f’’(c) > 0

f is concave down when…

f’’(c) < 0

inflection point of f

We say x=c is an inflection point of f if f’’ DNE or f’’ changes its sign at x=c.

d/dx sin x

cos x

d/dx cos x

-sin x

d/dx tan x

\sec^2x

d/dx cot x

-\csc^2x

d/dx sec x

\sec x\tan x

d/dx csc x

-\csc x\cot x

sinhx =

\frac{e^{x}-e^{-x}}{2}

coshx =

\frac{e^{x}+e^{-x}}{2}

tanhx =

\frac{\sinh x}{\cosh x}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}=\frac{e^{2x}-1}{e^{2x}+1}

(sinhx)’ =

coshx

(coshx)’ =

sinhx

(tanhx)’ =

\frac{1}{\cosh^2x}

\cosh^2x-\sinh^2x=

1