Created by Tynan Lazarus
November 15, 2014
1.1
Basic Derivatives
• Definitions of the Derivative:
f (x + h) − f (x)
h→0
h
d
•
(c) = 0
dx
lim
f (x) − f (a)
x→a
x−a
lim
•
d
(x) = 1
dx
•
d
(cx) = c
dx
•
d
(cu) = c · u0
dx
•
d n
(x ) = nxn−1 (Power Rule)
dx
•
d n
(u ) = nun−1 · u0 (Power Rule)
dx
•
d
(sin x) = cos(x)
dx
•
d
(sin u) = cos(u) · u0
dx
•
d
(cos x) = − sin(x)
dx
•
d
(cos u) = − sin(u) · u0
dx
•
d
(sec x) = sec(x) · tan(x)
dx
•
d
(sec u) = sec(u) · tan(u) · u0
dx
•
d
(csc x) = − csc(x) · cot(x)
dx
•
d
(csc u) = − csc(u) · cot(u) · u0
dx
•
d
(tan x) = sec2 (x)
dx
•
d
(tan u) = sec2 (u) · u0
dx
•
d
(cot x) = − csc2 (x)
dx
•
d
(cot u) = − csc2 (u) · u0
dx
•
d
1
(ln x) =
dx
x
•
d
1
(ln u) = · u0
dx
u
•
d x
(e ) = ex
dx
•
d u
(e ) = eu · u0
dx
•
d u
(a ) = au (ln a) · u0
dx
•
d
1
(loga u) =
· u0
dx
u · ln a
•
d
1
(arcsin u) = √
· u0
dx
1 − u2
•
d
−1
(arccos u) = √
· u0
dx
1 − u2
•
d
1
(arcsec u) = √
· u0
2
dx
|u| u − 1
•
d
−1
(arccsc u) = √
· u0
2
dx
|u| u − 1
•
1
d
(arctan u) =
· u0
2
dx
1+u
•
•
d
(u ± v) = u0 ± v 0
dx
•
d
(uv) = u0 v + uv 0 (Product Rule)
dx
a>0
a>0
−1
d
(arccot u) =
· u0
2
dx
1+u
d u u0 v − uv 0
•
=
(Quotient Rule)
dx v
v2
•
1
d
(f (g(x))) = f 0 (g(x)) · g 0 (x) (Chain Rule)
dx
Created by Tynan Lazarus
1.2
Facts About Derivatives
• Derivatives have many names. Problems may ask you to find
–
–
–
–
the
the
the
the
derivative at a point
slope of the tangent line at a point
rate of change at a point
velocity at a point
Each of the above ask for the exact same thing: to compute the derivative.
• When using the Chain Rule, its advisable to work from the outside function inwards. This
will be described in the Examples section.
• We need to take care of which variable with which we are taking a derivative. Notice that
d 2
5y = 0, since we are taking the derivative with respect to x. Since no x’s show up in the
dx
d x
equation, 5y 2 is treated as a constant. Similarly,
e + z = 0 + 1 = 1.
dz
• Position, velocity, acceleration and jerk are all linked through derivatives. If we denote position
as a function of time, s(t), velocity as v(t), acceleration as a(t), and jerk as j(t), then we have
the following relations:
ds
v(t) =
dt
1.3
a(t) =
dv
d2 s
= 2
dt
dt
j(t) =
d3 s
da
= 3
dt
dt
Examples
1. Find the derivative of ln (cos(4x + 2x2 )).
Solution: Using the Chain Rule, we start from the outside function, the natural log, and work
our way inwards to the cosine, then the polynomial. Thus,
d
1
d
ln cos(4x + 2x2 ) =
·
(cos(4x + 2x2 )
2
dx
cos(4x + 2x ) dx
1
d
=
· − sin(4x + 2x2 ) ·
4x + 2x2
2
cos(4x + 2x )
dx
1
=
· − sin(4x + 2x2 ) · (4 + 4x)
2
cos(4x + 2x )
sin(4x + 2x2 ) · (4 + 4x)
= −
cos(4x + 2x2 )
= −4 tan(4x + 2x2 )(1 + x).
2
Created by Tynan Lazarus
2. Find
d
sec arccot(6y 5 ) .
dy
Solution: Since we see that we need the Chain Rule, we start from the outside function,
secant, and work our way in through arc-cotangent to the polynomial. Thus,
d
d
sec arccot(6y 5 ) = sec arccot(6y 5 ) · tan arccot(6y 5 ) · arccot(6y 5 )
dy
dy
−1
d
= sec arccot(6y 5 ) · tan arccot(6y 5 ) ·
· 6y 5
5
2
1 + (6y ) dy
= sec arccot(6y 5 ) · tan arccot(6y 5 ) ·
13x
3. Find the slope of the tangent line of 2e
−1
· 30y 4
1 + (6y 5 )2
at x = 0.
Solution: First we need the general form of the derivative of our function. Notice that again
we need the Chain Rule. Thus,
d e13x
d 13x
13x
2
e
= 2e · ln(2) ·
dx
dx
d
13x
= 2e · ln(2) · e13x 13x
dx
e13x
13x
= 2
· ln(2) · e · 13
Now that we have the derivative, we can plug in x = 0 to find the derivative at our point.
Hence,
13(0)
0
2e
· ln(2) · e13(0) · 13 = 2e · ln(2) · e0 · 13 = 2 · ln(2) · 13 = 26 · ln(2).
4. Let v = x2 + y 2 , where x = cos(t) + sin(t) and y = cos(t) − sin(t). Find
dv
.
dt
Solution: There are two ways to do this problem. One requires the Chain Rule, the other just
requires some algebra. We will first show the algebra method.
Way 1: Instead of just trying to brute force the problem, we try to simplify our work for us.
If we make the substitutions for x and y, we see that
v = (cos(t) + sin(t))2 + (cos(t) − sin(t))2 .
FOILing out each piece, we see that
v = [cos2 (t) + 2 cos(t) sin(t) + sin2 (t)] + [cos2 (t) − 2 cos(t) sin(t) + sin2 (t)].
Notice that we can cancel the middle terms and combine the squared terms, and we are left
with
v = 2 cos2 (t) + 2 sin2 (t) = 2(cos2 (t) + sin2 (t)) = 2.
3
Created by Tynan Lazarus
So, taking the derivative is now very simple, since the derivative of a constant is 0. Thus,
dv
= 0.
dt
Way 2: Making the substitution, we see that
v = (cos(t) + sin(t))2 + (cos(t) − sin(t))2 .
Taking the derivative with respect to t, we use the Chain Rule on each set of parenthesis and
get that
dv
= [2(cos(t) + sin(t)) · (− sin(t) + cos(t))] + [2(cos(t) − sin(t)) · (− sin(t) − cos(t))].
dt
FOILing the terms, we get that
dv
= 2(cos2 (t)−sin2 (t))+2(− cos2 (t)+sin2 (t)) = 2 cos2 (t)−2 sin2 (t)−2 cos2 (t)+2 sin2 (t)) = 0.
dt
As we can see, both methods arrive at the same conclusion, that the derivative is equal to 0.
Its up to personal preference which way to choose. Usually it is more advisable to simplify
first and then take derivates, so Way 1 is more common. However, both methods are equally
valid.
5. Compute
d
ln(sin(x)) · arcsin(e2x ).
dx
Solution: Sometimes it is advisable to break up our functions into pieces to better keep track
of what we need to do. First, we notice that this is going to be a problem involving the Product
Rule. So, lets identify our f and g functions. Set f (x) = ln(sin(x)) and g(x) = arcsin(e2x ).
Since this is a product rule, we need f 0 (x) and g 0 (x). In both cases, we have to use the Chain
Rule. So, we have
1
· cos(x)
sin(x)
1
g 0 (x) = p
· e2x · 2
2x
2
1 − (e )
f 0 (x) =
f (x) = ln(sin(x))
g(x) = arcsin(e2x )
d
Recall that the Product Rule says
(f g) = f 0 g + f g 0 . Thus, all we need to do is insert
dx
everything into the right places. So we get
"
#
1
d
1
ln(sin(x)) · arcsin(e2x ) =
· cos(x) · arcsin(e2x ) + ln(sin(x)) · p
· e2x · 2
dx
sin(x)
1 − (e2x )2
2e2x
2x
= cot(x) · arcsin(e ) + ln(sin(x)) · √
1 − e4x
4