Calculus II, Chapt 7
Terms
undefined, object
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- Finding inverse of a function
-
1.) determine if function is one-to-one, consider domain
2.) write y in place of f(x)
3.) solve for x
4.) exchange x and y, consider domain
5.) replace y with f-1(x) - d/dx sin x
- cos x
- d/dx cos x
- -sin x
- d/dx tan x
- sec² x
- d/dx csc x
- -csc x cot x
- d/dx sec x
- sec x tan x
- d/dx cot x
- -csc² x
-
lim e^x
x->∞ - ∞
-
lim e^x
x->-∞ - 0
-
lim e^x-1/x
x->0 - 1
- eº =
- 1
- e¹ =
- e
- ln e =
- 1
- ln 1 =
- 0
- ln e³ =
- 3
- change of base
-
formula -
this -
equals this - ∫ x⿠dx
- (x^(n + 1) / n + 1) + C
- ∫ e^x dx
- e^x + C
- ∫ sin x dx
- -cos x + C
- ∫ sec² x dx
- tan x + C
- ∫ sec x tan x dx
- sec x + C
- ∫ sinh x dx
- cosh x + C
- ∫ tan x dx
- ln |sec x| + C
- ∫ 1 / x dx
- ln |x| + C
- ∫ a ^x dx
- (a^x / ln a) + C
- ∫ cos x dx
- sin x + C
- Theorem 7 differentiating inverse functions
- 1 / f'(f^-1)
- d/dx (ln |x|)
- 1 / x
- d/dx (arcsin x)
-
1 / √(1 - x²)
-1 < x < 1 - d/dx (arccos x)
-
-[ 1 / √(1 - x²) ]
-1 < x < 1 - d/dx (arctan x)
- 1 / 1 + x²
- d/dx (csc^-1 x)
- -[ 1 / (x)√(x² - 1) ]
- d/dx (sec^-1 x)
- 1 / (x)√(x² - 1)
- d/dx (cot^-1 x)
- -[ 1 / 1 + x² ]
- d/dx log base a (x)
- 1 / x ln(a)
- d/dx a^x
- a^x ln(a)
-
lim
x->0
(1 + x)^1/x - e
- arcsin(sin x)
- x for -π/2 ≦ x ≦ π/2
- sin (arcsin x)
- x for -1 ≦ x ≦ 1
- arccos(cos x)
- x for 0 ≦ x ≦ π
- cos (arccos x)
- x for -1 ≦ x ≦ 1