Calculus Formulas First derivative of a function f(x) =0 gives critical

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Calculus Formulas
First derivative of a function f(x) =0 gives critical points. Determines whether function increases or
decreases and determines extreme values. On a closed interval, check endpoints.
Second derivatives of a function f(x) gives point of inflections at changes in sign in the second derivative.
Sign pattern determines concave up /concave down.
Derivative forms
f '( x)  lim
f ( x  h)  f ( x )
f (a  h)  f (a)
; f '(a)  lim
h 0
h
h
f '(a)  lim
f ( x)  f (a)
f (a  h)  f (a  h)
; f '(a)  lim
h

0
xa
2h
h 0
x a
Derivative rules – note, if x is a function (u), use of chain rule is implied.
dy
c0
dx
dy n
u  nu n 1u '
dx
dy
1
special cases
x
dx
2 x
dy
uv  uv ' vu '
dx
dy u vu ' uv '

dx v
v2
,
dy 1 1

dx x x 2
dy
sin x  cos x
dx
dy
cos x   sin x
dx
dy
tan x  sec 2 x
dx
dy
sec x  sec x  tan x
dx
dy
csc x   csc x  cot x
dx
dy
cot x   csc 2 x
dx
dy
1
tan 1 x 
dx
1  x2
dy 1
1
sin x 
dx
1  x2
dy
sec 1 x 
dx
x
1
x2 1
inverse cot x, cos x, and csc x are each the opposite of theinverse tan x,sin x, and sec x
dy x
e  ex
dx
dy
1
ln x 
dx
x
dy x
a  a x ln a
dx
dy
1
log a x 
dx
x ln a
Generalized Chain Rule
d
( f ( g (x))  f '(g(x)) g'(x)
dx
Integral formulas
c  xC
x n 1
C
n 1
n
x 
e
x
 ex  C
1
 x  ln x  C
 sin kx 
 cos kx
C
k
 cos kx 
sin kx
C
k
 sec
2
kx 
tan kx
C
k
Average Value of any function on [a,b]
b
1
f ( x) dx
b  a a
Extreme Value Theorem
On a closed continuous interval [a,b], there must be at least one max and one min value.
Intermediate Value Theorem
On a closed continuous interval [a,b], every value f(c) must be between f(a) and f(b) for all c on [a,b].
- This theorem applies for derivatives as well
Mean Value Theorem
On closed continuous [a,b], the slope of the secant line must equal the slope of the tangent line at least
once. f '(c) 
f (b)  f (a)
if c on (a, b)
ba
FTC
x
1)
dy
f (t )dt  f ( x ) , note if the upper limit of integration is a function of x, you must multiply
dx a
f(x) by the derivative of the upper limit of integration
dy
dx
g ( x)

f (t )dt  f (g( x))  g'(x)
a
b
2)
 f (x) dx  F (b)  F(a) if F(x) is any antiderivative of f.
a
Integral Approximations
RRAM, LRAM, MRAM – estimates of areas under curves
Underestimate if below curve, upper estimate if above curve
Trapezoidal Approximations
A=
1
h
h(b1  b2 ) or trapezoidal rule ( y0  2 y1  2 y2  .....  2 yn 1  yn )
2
2
Integral Techniques
U-substitution including changing of limits of integration
Separation of variables
Position Velocity and Acceleration * do not forget initial conditions
Derivative of position is velocity
Derivative of velocity is acceleration
Integral of acceleration is velocity
Integral of velocity is position
Absolute value of velocity is speed.
Displacement is net distance. Total distance is integral absolute value velocity.
Inverses- If f(g(x))=x then you have an inverse. An inverse is when x and y are switched in an order pair.
If you are trying to find the derivative of an inverse at a point (a,b), it is
tangent line to the inverse at (b,a) is y  a 
1
( x  b)
f '(a)
1
and the equation of the
f '( a )
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