Derivatives
Function
xn
ex
ax
uv
Derivative
nx n-1
ex
(ln a)ax
dv
du
u v
dx
dx
u
v
du
dv
u
dx
dx
v2
dy dy du


dx du dx
If y = f(u) and
u = g(x), then
sin x
cos x
tan x
ln x *
arctan x *
arcsin x *
v
cos x
- sin x
1
cos 2 x
1
x
1
1 x2
1
1 x2
*Derived on pgs. 226-27 of Hughes-Hallett
Integrals
Function
xn
Integral
ex
sin x
cos x
tan x
ex
- cos x
sin x
- ln | cos x |
tan x
1
cos 2 x
1
x
dv
 u dx dx
ln x
ax
x n 1
n 1
ln x
uv   v
du
dx
dx
x ln x – x , x > 0
1 x
a
ln a
Implicit Differentiation
To implicitly differentiate an expression of the form, (say),
x 2  xy  y 2  8 , we take
Where
d 2
d
d
d
( x )  ( xy)  ( y 2 ) 
(8)
dx
dx
dx
dx
d 2
d 2 dy
dy
(y ) 
(y )
 2y
dx
dy
dx
dx
and we use the product rule for
d
(xy) to get:
dx
d
d
dy
dy
( x)  y  ( y )  x  y  x
dx
dy
dx
dx
We then “solve for”
dy
:
dx
d 2
d
d
d
( x )  ( xy)  ( y 2 ) 
(8)
dx
dx
dx
dx
2x  y  x
dy
dy
 2y
0
dx
dx
dy
( x  2 y )  2 x  y
dx
dy  2 x  y
(Then one can calculate the value of the derivative at a given point by plugging

dx
x  2y
in the x and y values at that point.)
We may also apply the chain rule to implicit differentiation:
Ex. sin(xy) = 2x + 5
d
d
d
sin( xy) 
2x  5
dx
dx
dx
Note that y = sin u and u = xy, so
so
x cos( xy)
dy
du
dy
 cos u  cos( xy) ,
x
 y,
du
dx
dx
dy
 dy

 cos( xy)  x
 y  . Returning to the above implicit differentiation:
dx
 dx

dy
 y cos( xy)  2  0
dx
dy 2  y cos( xy)

dx
x cos( xy)
Integration Methods
1. Substitution
dw
, will “cancel out” other terms. We then
dx
dw
dx .
f ( x)dx into the form  f ( w)
dx
Let w be an “inside function” whose derivative,
reassemble an integral of the form
Ex.
 3x
2
cos( x 3 )dx

Let w = x3 , then
 cos( w)dw  sin w  C  sin( x
3
dw
 3x 2 and dw = 3x2dx
dx
)C
2. Integration by Parts (Formula is shown in anti-derivative table)
Ex.
x
3
u = ln x
ln( x)dx
du 1

dx x
x
3
ln( x)dx 
dv
 x3
dx
x4
v
4
1 4
1
1
1
x ln x   x 3  x 4 ln x  x 4  C
4
4
4
16
Note: Often when an integral is in the form
not always, as shown in the example above)
3. Partial Fractions
Ex.
1
 ( x  2)( x  5) dx
1
A
B


( x  2)( x  5) x  2 x  5
1 = A(x – 5) + B(x - 2)
Let x = 5, then 1 = 3B , so B = 1/3
Let x = 2, then 1 = - 3A, so A = - 1/3.
∫ xa f(x) dx , it is useful to let u = xa. (Certainly
So,
1
1
1
1
1
1
1
 ( x  2)( x  5) dx   3  x  2 dx  3  x  5 dx   3 ln | x  2 |  3 ln | x  5 | C
Note: If the highest power of the term in the numerator is greater than or equal to the term in
the denominator, you will likely need to perform a division, get a remainder, and express the
fraction as a non-fraction plus a fractional denominator.
4. Estimating the Integral Function when the Anti-Derivative Cannot be Found
Algebraically
If an integral [ F(x) ] cannot be found algebraically, we can still find values of the integral
function by estimating the area under the curve of f(x). [where F′(x) = f(x) ].
x
sin t
dt
t
0
Ex. F ( x)  
0
1
2
3
x
F(x) 0 0.95 1.61 1.85
x
Ex. If the function g defined by g ( x)   sin( t 2 )dt on the closed interval - 1 ≤ x ≤ 3, then g
0
has a local minimum at x = ?
Note that for the domain - 1 ≤ x ≤ 3, the area is positive between – 1 and 1.8, then some area is
“subtracted off”. The location where the most area has been “subtracted off” corresponds to the
local minimum of g.
Likewise, we can see that the maximum value of g on the interval - 1 ≤ x ≤ 3 must occur at
approximately x = 1.8.
 e x
Ex. If f ( x)  sin 
 2

 and f(0) = 1, then what is f(2)?

In this type of estimation, we normally assume that the y-intercept occurs at the origin. Since
f(0) = 1 in this case, we know that all values of f have been “pushed up” by 1.
 e x 
 on the calculator from 2 to 0, the calculator
So, in calculating the area of f ( x)  sin 
2


gives the value 0.16, so the solution for f(2) = 0.16 + 1 ≈ 1.16.
Solving First-Order Differential Equations
This will be an equation in which we are shown the derivative function, and be asked to
determine the initial function. Obviously, the solution to any given differential equation will be
a whole family of functions (each with a different constant), but if we are given some initial
condition, we can solve for the particular equation.
To solve this, we simply “gather” the y’s on one side and the x’s on the other, then integrate both
sides. Often, we will want to then isolate the y.
Ex.
dy
 ky
dx
e ln y  e kx C
1
dy  (k )dx
y
1
 y dy   (k )dx
ln y  kx  C
y  e kx e C , but eC is just some constant, so let A = eC ,
then, y  Ae kx
Ex.
dy x 3

Determine the solution to the differential equation
if y(2) = 3.
dx y 2
y 2 dy  x 3 dx
2
3
 y dy   x dx
y3 x4

 C , but when x = 2, y = 3, so
3
4
27 16

C
3
4
9 = 4 + C , so C = 5:
y3 x4

5
3
4
y3 
3 4
x  15
4
y3
3 4
x  15
4