'Derivatives' Algebra Reference Sheet
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1Derivatives
Surfaces
1
2
sec(tan
(x)) =
1 + x
Projection of ~ u onto ~ v :
Given z=f(x,y), the partial derivative of
~ u ·~ v
1
pr
~ u = (
)~ v
z with respect to x is:
x
x
tan(sec
(x))
D
e
= e
~ v
Ellipsoid
||~ v || 2
x
@f (x,y)
@z
2
f
(x, y) = z
=
=
D
sin(x) = cos(x)
= ( x
1 if x
1)
x 2
y 2
z 2
x
x
x
@x
@x
+
+
= 1
Cross Product
a 2
b 2
c 2
likewise for partial with respect to y:
D
cos(x) =
sin(x)
= (
x
2
1 if x
1)
x
2
~ u
~ v
@f (x,y)
@z
D
tan(x) = sec
(x)
1
f
(x, y) = z
=
=
2
x
sinh
(x) = ln x +
x
+ 1
y
y
@y
@y
2
Produces a Vector
D
cot(x) =
csc
(x)
1
Notation
x
sinh
(x) = ln x +
x
2
1, x
1
(Geometrically, the cross product is the
D
sec(x) = sec(x) tan(x)
For f
, work ”inside to outside” f
x
1
1+x
1
tanh
(x) =
ln x +
, 1 < x <
1
xyy
x
area of a paralellogram with sides ||~ u ||
D
csc(x) =
csc(x) cot(x)
2
1
x
x
then f
, then f
xy
xyy
and ||~ v ||)
1
x 2
1
1+
1
D
sin
=
, x
[ 1, 1]
1
sech
(x) = ln[
], 0 < x
1
@ 3 f
x
Hyperboloid of One Sheet
f
=
,
x 2
x
~ u =< u
, u
, u
>
1
xyy
1
2
3
@x@ 2 y
e x
x
y 2
1
1
e
x 2
z 2
sinh(x) =
D
cos
=
, x
[ 1, 1]
~ v =< v
, v
, v
>
+
= 1
@ 3 f
1
2
3
x
2
a 2
b 2
c 2
x 2
For
, work right to left in the
1
e x +e
x
@x@ 2 y
(Major Axis: z because it follows - )
⇡
cosh(x) =
1
1
⇡
D
tan
=
,
x
ˆ k
ˆ i
ˆ j
2
denominator
x
1+x 2
2
2
~ u
~ v =
1
1
u
u
u
D
sec
=
, |x| > 1
Trig Identities
1
2
3
x
Gradients
x 2
|x|
1
v
v
v
1
2
3
D
sinh(x) = cosh(x)
2
2
The Gradient of a function in 2 variables
x
sin
(x) + cos
(x) = 1
D
cosh(x) = sinh(x)
~ v = ~ 0 means the vectors are paralell
~ u
2
2
is
f =< f
, f
>
x
1 + tan
(x) = sec
(x)
x
y
2
D
tanh(x) = sech
(x)
2
2
The Gradient of a function in 3 variables
x
1 + cot
(x) = csc
(x)
2
Lines and Planes
D
coth(x) =
csch
(x)
sin(x ± y) = sin(x) cos(y) ± cos(x) sin(y)
is
f =< f
, f
, f
>
x
x
y
z
Hyperboloid of Two Sheets
D
sech(x) =
sech(x) tanh(x)
x
cos(x ± y) = cos(x) cos(y) ± sin(x) sin(y)
Equation of a Plane
z 2
x 2
y 2
Chain Rule(s)
D
csch(x) =
csch(x) coth(x)
= 1
x
tan(x)±tan(y)
(x
, y
, z
) is a point on the plane and
c 2
a 2
b 2
tan(x ± y) =
0
0
0
1
1
D
sinh
=
1
tan(x) tan(y)
(Major Axis: Z because it is the one not
Take the Partial derivative with respect
< A, B, C > is a normal vector
x
x 2 +1
sin(2x) = 2 sin(x) cos(x)
subtracted)
to the first-order variables of the
1
1
2
2
D
cosh
=
, x > 1
cos(2x) = cos
(x)
sin
(x)
x
function times the partial (or normal)
x 2
A(x
x
) + B(y
y
) + C(z
z
) = 0
1
2
2
cosh(n
x)
sinh
x = 1
0
0
0
1
derivative of the first-order variable to
1
D
tanh
=
1 < x < 1
< A, B, C > · < x x
, y y
, z z
>= 0
2
2
x
1 + tan
(x) = sec
(x)
0
0
0
x 2
1
the ultimate variable you are looking for
Ax + By + Cz = D where
2
2
1
1
1 + cot
(x) = csc
(x)
D
sech
=
, 0 < x < 1
summed with the same process for other
x
x 2
D = Ax
+ By
+ Cz
x
1
1
cos(2x)
0
0
0
2
sin
(x) =
first-order variables this makes sense for.
1
D
ln(x) =
2
x
x
1+cos(2x)
2
Example:
cos
(x) =
Equation of a line
2
let x = x(s,t), y = y(t) and z = z(x,y).
Integrals
Elliptic Paraboloid
2
1
cos(2x)
A line requires a Direction Vector
tan
(x) =
1+cos(2x)
y 2
z then has first partial derivative:
x 2
~ u =< u
, u
, u
> and a point
z =
+
1
2
3
sin( x) =
sin(x)
1
@z
@z
dx = ln |x| + c
a 2
b 2
and
(x
, y
, z
)
x
@x
@y
cos( x) = cos(x)
1
1
1
(Major Axis: z because it is the variable
x
x
e
dx = e
+ c
x has the partial derivatives:
then,
tan( x) =
tan(x)
NOT squared)
@x
@x
x
1
x
and
a
dx =
a
+ c
a parameterization of a line could be:
@s
@t
ln a
and y has the derivative:
ax
ax
1
x = u
t + x
e
dx =
e
+ c
Calculus 3 Concepts
1
1
a
dy
y = u
t + y
1
1
dx = sin
(x) + c
2
1
dt
x 2
In this case (with z containing x and y
Cartesian coords in 3D
z = u
t + z
1
3
1
1
1
as well as x and y both containing s and
dx = tan
(x) + c
1+x 2
given two points:
@z
@z
@z
@x
t), the chain rule for
is
=
Distance from a Point to a Plane
1
1
dx = sec
(x) + c
@s
@s
@x
@s
(x
, y
, z
) and (x
, y
, z
),
1
1
1
2
2
2
@z
x 2
The distance from a point (x
, y
, z
) to
The chain rule for
is
x
1
0
0
0
Hyperbolic Paraboloid
Distance between them:
@t
sinh(x)dx = cosh(x) + c
a plane Ax+By+Cz=D can be expressed
dy
@z
@z
@x
@z
(Major Axis: Z axis because it is not
=
+
(x
x
)
2
+ (y
y
)
2
+ (z
z
)
2
1
2
1
2
1
2
@t
@x
@t
@y
dt
cosh(x)dx = sinh(x) + c
by the formula:
squared)
Midpoint:
Note: the use of ”d” instead of ”@” with
|Ax 0 +By 0 +Cz 0
tanh(x)dx = ln | cosh(x)| + c
D|
d =
y 2
x 1 +x 2
y 1 +y 2
z 1 +z 2
x 2
the function of only one independent
(
,
,
)
z =
A 2 +B 2 +C 2
tanh(x)sech(x)dx =
sech(x) + c
2
2
2
b 2
a 2
variable
Sphere with center (h,k,l) and radius r:
2
sech
(x)dx = tanh(x) + c
2
2
2
2
(x
h)
+ (y
k)
+ (z
l)
= r
Coord Sys Conv
Limits and Continuity
csch(x) coth(x)dx =
csch(x) + c
ln | cos(x)| + c
tan(x)dx =
Vectors
Limits in 2 or more variables
Cylindrical to Rectangular
cot(x)dx = ln | sin(x)| + c
x = r cos(✓)
Limits taken over a vectorized limit just
Vector: ~ u
cos(x)dx = sin(x) + c
Elliptic Cone
evaluate separately for each component
y = r sin(✓)
Unit Vector: ˆ u
sin(x)dx =
cos(x) + c
(Major Axis: Z axis because it’s the only
of the limit.
z = z
2
2
2
Magnitude: ||~ u || =
u
+ u
+ u
1
1
u
one being subtracted)
dx = sin
(
) + c
1
2
3
Rectangular to Cylindrical
Strategies to show limit exists
a
a 2
u 2
y 2
~ u
x 2
z 2
Unit Vector: ˆ u =
1. Plug in Numbers, Everything is Fine
2
2
+
= 0
r =
x
+ y
1
1
1 u
dx =
tan
+ c
||~ u ||
a 2
b 2
c 2
a 2 +u 2
a
a
y
2. Algebraic Manipulation
tan(✓) =
x
ln(x)dx = (xln(x))
x + c
Dot Product
-factoring/dividing out
z = z
~ u · ~ v
-use trig identites
Spherical to Rectangular
U-Substitution
Produces a Scalar
3. Change to polar coords
x = ⇢ sin( ) cos(✓)
Let u = f (x) (can be more than one
(Geometrically, the dot product is a
if (x, y)
(0, 0)
r
0
y = ⇢ sin( ) sin(✓)
variable).
vector projection)
Strategies to show limit DNE
z = ⇢ cos( )
f (x)
~ u =< u
, u
, u
>
1. Show limit is di↵erent if approached
Determine: du =
dx and solve for
Rectangular to Spherical
1
2
3
Cylinder
dx
~ v =< v
, v
, v
>
from di↵erent paths
dx.
1
2
3
⇢ =
x
2
+ y
2
+ z
2
1 of the variables is missing
~ u · ~ v = ~ 0 means the two vectors are
2
(x=y, x = y
, etc.)
y
Then, if a definite integral, substitute
tan(✓) =
OR
x
Perpendicular ✓ is the angle between
2. Switch to Polar coords and show the
the bounds for u = f (x) at each bounds
z
2
2
cos( ) =
(x
a)
+ (y
b
) = c
them.
x 2 +y 2 +z 2
limit DNE.
Solve the integral using u.
(Major Axis is missing variable)
~ u · ~ v = ||~ u || ||~ v || cos(✓)
Spherical to Cylindrical
Continunity
r = ⇢ sin( )
~ u · ~ v = u
v
+ u
v
+ u
v
A fn, z = f (x, y), is continuous at (a,b)
Integration by Parts
1
1
2
2
3
3
Partial Derivatives
✓ = ✓
NOTE:
if
udv = uv
vdu
z = ⇢ cos( )
ˆ u · ˆ v = cos(✓)
Partial Derivatives are simply holding all
f (a, b) = lim
f (x, y)
(x,y)
(a,b)
2
Cylindrical to Spherical
||~ u ||
= ~ u · ~ u
other variables constant (and act like
Which means:
Fns and Identities
⇢ =
r
2
+ z
2
~ u · ~ v = 0 when
constants for the derivative) and only
1. The limit exists
1
sin(cos
(x)) =
1
x
2
✓ = ✓
Angle Between ~ u and ~ v :
taking the derivative with respect to a
2. The fn value is defined
z
1
1
~ u ·~ v
2
cos( ) =
cos(sin
(x)) =
1
x
✓ = cos
(
)
given variable.
3. They are the same value
||~ u || ||~ v ||
r 2 +z 2
Directional Derivatives
Double Integrals
Surface Integrals
Other Information
Work
Let ~ F = M ˆ i + ˆ j + ˆ k (force)
a
a
With Respect to the xy-axis, if taking an
Let
=
Let z=f(x,y) be a fuction, (a,b) ap point
b
b
M = M (x, y, z), N = N (x, y, z), P =
integral,
·R be closed, bounded region in xy-plane
in the domain (a valid input point) and
Where a Cone is defined as
P (x, y, z)
·f be a fn with first order partial
dydx is cutting in vertical rectangles,
ˆ u a unit vector (2D).
z =
a(x
2
+ y
2
),
(Literally)d~ r = dx ˆ i + dy ˆ j + dz ˆ k
derivatives on R
dxdy is cutting in horizontal
The Directional Derivative is then the
In Spherical Coordinates,
rectangles
~ F · d~ r
·G be a surface over R given by
derivative at the point (a,b) in the
Work w =
1
a
= cos
(
)
c
z = f (x, y)
1+a
direction of ˆ u or:
(Work done by moving a particle over
Right Circular Cylinder:
·g(x, y, z) = g(x, y, f (x, y)) is cont. on R
Polar Coordinates
D
f (a, b) = ˆ u ·
f (a, b)
curve C with force ~ F )
~ u
2
2
V = ⇡r
h, SA = ⇡r
+ 2⇡rh
Then,
This will return a scalar. 4-D version:
When using polar coordinates,
pn
mp
m
lim
(1 +
)
= e
g(x, y, z)dS =
f (a, b, c) = ˆ u ·
dA = rdrd✓
D
f (a, b, c)
n
inf
n
~ u
G
Independence of Path
Law of Cosines:
g(x, y, f (x, y))dS
R
2
2
2
Surface Area of a Curve
a
= b
+ c
2bc(cos(✓))
Tangent Planes
2
2
where dS =
f
+ f
+ 1dydx
Fund Thm of Line Integrals
x
y
let z = f(x,y) be continuous over S (a
Stokes Theorem
let F(x,y,z) = k be a surface and P =
C is curve given by ~ r (t), t
[a, b];
Flux of ~ F across G
closed Region in 2D domain)
(x
, y
, z
) be a point on that surface.
~ r (t) exists. If f (~ r ) is continuously
~ F · ndS =
Let:
0
0
0
Then the surface area of z = f(x,y) over
Equation of a Tangent Plane:
di↵erentiable on an open set containing
G
·S be a 3D surface
S is:
[ M f
N f
+ P ]dxdy
F (x
, y
, z
)· < x
x
, y
y
, z
z
>
f (~ r ) · d~ r = f ( ~ b)
x
y
· ~ F (x, y, z) =
R
0
0
0
0
0
0
C, then
f (~ a )
2
2
where:
SA =
f
+ f
+ 1dA
c
S
x
y
Equivalent Conditions
M (x, y, z) ˆ i + N (x, y, z) ˆ j + P (x, y, z) ˆ l
· ~ F (x, y, z) =
Approximations
~ F (~ r ) continuous on open connected set
st
Triple Integrals
·M,N,P have continuous 1
order partial
M (x, y, z) ˆ i + N (x, y, z) ˆ j + P (x, y, z) ˆ k
D. Then,
derivatives
let z = f (x, y) be a di↵erentiable
·G is surface f(x,y)=z
f (x, y, z)dv =
(a) ~ F =
f for some fn f. (if ~ F is
·C is piece-wise smooth, simple, closed,
function total di↵erential of f = dz
s
·~ n is upward unit normal on G.
2 (x)
2 (x,y)
a 2
f (x, y, z)dzdydx
dz =
f · < dx, dy >
curve, positively oriented
conservative)
st
·f(x,y) has continuous 1
order partial
a 1
1 (x)
1 (x,y)
· ˆ T is unit tangent vector to C.
This is the approximate change in z
~ F (~ r ) · d~ r isindep.of pathinD
Note: dv can be exchanged for dxdydz in
(b)
derivatives
c
Then,
The actual change in z is the di↵erence
any order, but you must then choose
~ F (~ r ) · d~ r = 0 for all closed paths
(c)
~ F
· ˆ T dS =
~ F ) · ˆ n dS =
in z values:
your limits of integration according to
c
(
c
in D.
s
z = z
z
that order
~ F ) · ~ n dxdy
1
(
Conservation Theorem
R
~ F = M ˆ i + N ˆ j + P ˆ k continuously
Remember:
Unit Circle
Maxima and Minima
Jacobian Method
~ F · ~ T ds =
(M dx + N dy + P dz)
di↵erentiable on open, simply connected
c
f (g(u, v), h(u, v))|J(u, v)|dudv =
(cos, sin)
Internal Points
set D.
G
~ F conservative
~ F = ~ 0
1. Take the Partial Derivatives with
f (x, y)dxdy
R
~ F = ~ 0 i↵ M
respect to X and Y (f
and f
) (Can use
(in 2D
= N
)
x
y
y
x
@x
@x
gradient)
J(u, v) =
@u
@v
@y
@y
2. Set derivatives equal to 0 and use to
@u
@v
Green’s Theorem
solve system of equations for x and y
Common Jacobians:
3. Plug back into original equation for z.
(method of changing line integral for
Rect. to Cylindrical: r
Use Second Derivative Test for whether
double integral - Use for Flux and
2
Rect. to Spherical: ⇢
sin( )
points are local max, min, or saddle
Circulation across 2D curve and line
Vector Fields
integrals over a closed boundary)
Second Partial Derivative Test
M dy
N dx =
(M
+ N
)dxdy
x
y
R
1. Find all (x,y) points such that
let f (x, y, z) be a scalar field and
M dx + N dy =
(N
M
)dxdy
x
y
~ F (x, y, z) =
f (x, y) = ~ 0
R
Let:
2
M (x, y, z) ˆ i + N (x, y, z) ˆ j + P (x, y, z) ˆ k be
2. Let D = f
(x, y)f
(x, y)
f
(x, y)
xx
yy
·R be a region in xy-plane
xy
IF (a) D > 0 AND f
< 0, f(x,y) is
a vector field,
·C is simple, closed curve enclosing R
xx
@f
@f
@f
local max value
Grandient of f =
f =<
,
,
>
(w/ paramerization ~ r (t))
@x
@y
@z
(b) D > 0 AND f
(x, y) > 0 f(x,y) is
· ~ F (x, y) = M (x, y) ˆ i + N (x, y) ˆ j be
xx
Divergence of ~ F :
local min value
· ~ F =
continuously di↵erentiable over R C.
@M
@N
@P
+
+
(c) D < 0, (x,y,f(x,y)) is a saddle point
@x
@y
@z
Form 1: Flux Across Boundary
Curl of ~ F :
(d) D = 0, test is inconclusive
~ n = unit normal vector to C
ˆ i
ˆ j
ˆ k
3. Determine if any boundary point
~ F · ~ n =
· ~ F dA
~ F =
@
@
@
c
R
gives min or max. Typically, we have to
@x
@y
@z
M dy
N dx =
(M
+ N
)dxdy
x
y
parametrize boundary and then reduce
R
M
N
P
Form 2: Circulation Along
to a Calc 1 type of min/max problem to
Boundary
Line Integrals
solve.
~ F · d~ r =
~ F · ˆ u dA
The following only apply only if a
C given by x = x(t), y = y(t), t
[a, b]
c
R
M dx + N dy =
(N
M
)dxdy
boundary is given
b
x
y
f (x, y)ds =
f (x(t), y(t))ds
R
c
a
Area of R
1. check the corner points
dy
1
1
dx
2
2
A =
(
ydx +
xdy)
where ds =
(
)
+ (
)
dt
2. Check each line (0
x
5 would
2
2
dt
dt
give x=0 and x=5 )
dy
or
1 + (
)
2
dx
On Bounded Equations, this is the
dx
Gauss’ Divergence Thm
global min and max...second derivative
dx
or
1 + (
)
2
dy
dy
test is not needed.
(3D Analog of Green’s Theorem - Use
To evaluate a Line Integral,
for Flux over a 3D surface) Let:
· get a paramaterized version of the line
Lagrange Multipliers
· ~ F (x, y, z) be vector field continuously
(usually in terms of t, though in
di↵erentiable in solid S
Given a function f(x,y) with a constraint
exclusive terms of x or y is ok)
·S is a 3D solid ·@S boundary of S (A
g(x,y), solve the following system of
· evaluate for the derivatives needed
Originally Written By Daniel Kenner for
Surface)
equations to find the max and min
(usually dy, dx, and/or dt)
MATH 2210 at the University of Utah.
·ˆ n unit outer normal to @S
points on the constraint (NOTE: may
· plug in to original equation to get in
Source code available at
Then,
need to also find internal points.):
terms of the independant variable
CheatSheet
~ F (x, y, z) · ˆ n dS =
· ~ F dV
f =
g
· solve integral
Thanks to Kelly Macarthur for Teaching and
@S
S
g(x, y) = 0(orkif given)
(dV = dxdydz)
Providing Notes.
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