Unit 6 Flashcards
a function is continuous where it is differentiable
definition of a derivative
f’(a) = lim f(a+h)-f(a) / h
h->0
implicit differentiation
technique we use to find a derivative when y is not defined explicitly in terms of x but is differentiable
derivative of the inverse of a function
(f^-1)’(x) = 1 / f’(f^-1(x))
DERIVATIVES OF INVERSES
d/dx arcsin(u) =
1 / (1 - u^2)^1/2 du/dx
d/dx arccos(u) =
- d/dx arcsin(u)
d/dx arctan(u) =
1 / 1 + u^2 du/dx
d/dx arccot(u) =
- d/dx arctan(u)
d/dx arcsec(u) =
1 / |u|(u^2 - 1)^1/2 du/dx
d/dx arcsec(u) =
- d/dx arcsec(u)
∫ 1/u du =
ln|u| + C
∫ a^u du =
a^u / ln(a) + C
mean value theorem
- f(x) is continuous on [ , ] and differentiable on ( , ), therefore the mean value theorem can be applied
- f(b)-f(a) / b-a = f’(C)
- solve for C
average rate
slope of endpoints: f(b)-f(a) / b-a
average value
. b
1/b-a ∫ f(x) dx
. a
average rate = average value, it just depends on _______
what information you are being given
instant rate
derivative at the given point
area between curves
in terms of x:
∫ (top function - bottom function)dx
in terms of y:
∫(right function - left function)dy
Volume Disk method
π ∫ [r(x)²] dx
- needs to be in terms of what axis it is being rotated about
Volume Washer method
π ∫ (R² - r²)dx
volume by known cross section
= ∫ area of shape dx
- will be given b, plug into area equation
