Download Higher order derivatives

Higher order derivatives
Definition. If the derivative of f, that is f ′, is itself
differentiable at a point x, we call the result the second
derivative of f and denote it by the various notations:
2
2y
d
d
f ′′(x)
where y = f(x)
 f ( x) 
dx 2
dx2
The derivative of the second derivative is the third
derivative, and so on. In general the result of taking the
derivative n times is called the nth derivative of f with
respect to x and is denoted variously as:
n primes
f '' '( x)
d n f (x)


dx n
dny
dxn
Example. Find the third derivative of f(x) = 3x5
Solution. f ′(x )=15x4
f ′′(x) = 60 x3
f ′′′(x) =180x2
Example. Find the second derivative of f(x) =
Solution.







2
′







3
2
2
x
(1
+
x
)(2
x
)
−
x
(3
x
)
f ′(x) =
=
1+ x3
(1+ x3) 2
4
2
x
−
x
=
(1+ x3) 2
x2
1+ x3
Thus
′
′
f ′′(x) = 2 x − x = (1+ x ) (2 x − x )′− (2x − x )((1+ x ) )
(1+ x3)2
(1+ x3)4







4







3 2
4
4
3 2
3
4
2
5
(1
+
x
)
(2
−
4
x
)
−
(2
x
−
x
)(6
x
+
6
x
)
=
(1+ x3)4
3 2
3
3
3
3
(1
+
x
)
(2
−
4
x
)
−
(2
−
x
)6
x
(1
+
x
)
=
(1+ x3)4
3
3
3
3
(1
+
x
)(2
−
4
x
)
−
(2
−
x
)6
x
=
(1+ x3)3
3
6
2
−
14
x
+
2
x
=
(1+ x3)3
3 2
Derivatives of the Trigonometric Functions
Recall two important limits that involve the trigonometric
functions sine and cosine.
lim sin( x) =1
x →0 x
lim 1−cos(x) = 0
x
x →0
The one other trigonometric identities that we use are:
sin(α + β ) =sin(α )cos(β ) + cos(α )sin(β )
cos(α + β ) = cos(α )cos(β ) −sin(α )sin(β )
It is worth noting here that virtually all trigonometric identities
used in calculus are derived from these three fundamental
trigonometric identities:
sin2 (α ) + cos2 (α ) =1
sin(α + β ) =sin(α )cos(β ) + cos(α )sin(β )
cos(α + β ) = cos(α )cos(β ) −sin(α )sin(β )
The derivative of the sine function is computed as follows:
sin(x)′ = lim sin(x + h) −sin( x)
h
h →0
= lim sin( x)cos(h) + cos( x)sin(h) −sin(x)
h
h →0
 cos(h)−1 
 sin(h) 
= lim sin( x) 
+
cos(
x
)

 h 
h
h →0




 cos(h)−1 
 sin(h) 
= sin(x) lim 
+ cos( x) lim 


h
h
h →0 
h
→
0



= cos( x)
Thus we have shown that
d sin( x) = cos( x) or equivalently sin(x) ′ = cos( x)








dx
The derivative of the cosine function is found in almost
identical way, using another of the fundamental identities
d cos( x) = lim cos( x + h) −cos(x)


dx
h
h →0
= lim [cos( x)cos(h) −sin( x)sin(h)]−cos(x)
h
h →0
= lim cos(x)[cos(h)−1] − sin( x)sin(h)
h
h
h →0
= lim cos(x)[cos(h)−1] − lim sin(x)sin(h)
h
h
h →0
h →0
= cos( x) lim [cos(h)−1] −sin(x) lim sin(h)
h
h →0
h →0 h
=−sin( x)
Thus we have shown that
d cos(x) =−sin(x) or equivalently cos(x) ′ =−sin(x)








dx
The derivatives of the other trigonometric functions can be
found by using the quotient rule for differentiation.
Theorem.
(tan( x))′ = sec2( x)
(sec(x))′ = sec( x)tan( x)
(cot( x))′ =−csc2( x)
(csc(x))′ =−csc( x)cot( x)
As an example we show the result for tan(x) and sec(x).
 sin( x)
′
(tan( x)) = 
 cos(x)
′ cos(x)sin(x)′−sin( x)cos(x)′
=

cos2( x)
= cos(x)cos( x )−sin( x)(− sin(x))
cos 2 ( x)
2
2
cos
(
x
)
+
sin
( x) =
1 =sec2 (x)
=
cos 2 (x)
cos 2 (x)
Similarly we have
 1
′
(sec(x)) = 
 cos( x)
′ cos( x)(1)′ −1cos( x)′ sin( x)
=
=

cos 2 ( x)
cos 2 (x)
 1  sin(x) 
=
= sec( x)tan( x)


 cos( x)  cos( x) 
We summarize the results.
d sin(x) = cos( x)


dx
d cos(x) =−sin( x)


dx
d tan(x) = sec2( x)


dx
d cot( x) =−csc2( x)


dx
d sec( x) = sec(x)tan(x)


dx
d csc( x) =−csc( x)cot(x)


dx
With these results, we can now combine trigonometric
functions with the other functions that we can differentiate,
to form a larger class of problems that we can do.
Example. Find the derivative of the function
f ( x) = 7cos(x ) − 2tan(x )
Solution. f ′(x) = 7 cos(x) ′ − 2 tan(x) ′
(
)
(
(
) (
= 7 − sin( x) − 2 sec2( x)
)
)
=−7sin( x) − 2sec2( x)
Example. Find the derivative of the function
f ( x) = cos( x)
x sin(x)
Solution.
′− (cos(x))(x sin( x))′
(
x
sin(
x
))(cos(
x
))
′
f (x) =
(x sin(x)) 2
Let us interrupt this computation by noting that the product
formula lets us compute the derivative of the function x sin(x)
The result is:
(xsin( x))′ = xcos(x) +sin(x)
Returning to the main computation, we have
′
′
f ′(x) = ( x sin(x))(cos( x)) − (cos(x))(x sin( x))
(x sin(x)) 2
= (x sin( x))(− sin(x)) − (cos(x))(x cos(x) + sin( x))
(x sin( x)) 2
2
2
−
x
sin
(
x
)
−
x
cos
(x )−cos(x)sin(x))
=
(x sin( x))2
= − x −cos(x)sin(x))
( x sin(x)) 2
Example. Find the second derivative of the function
f ( x) = x2 cos(x)
Solution: By the product formula, we have
f ′(x) = (2 x)cos( x) − x2 sin( x)
Using the product formula twice more gives us the second
derivative.
′
2
′
f ′′(x) = 2x cos( x) − x sin( x)
(
)
(
)
= (2cos( x) − 2 xsin(x)) − (2 x)sin( x) − x2 cos( x)
= 2cos(x) − 4 xsin(x)) − x2 cos( x)
Example. An airplane is flying on a horizontal path at a height
of 3800ft, as shown in the figure. At what rate is the distance s
between the airplane and the fixed point P changing with
respect to θ when θ = 30° ?
s
P
θ
3800 ft.
Solution. It follows from basic trigonometry that
csc(θ ) = s or s = 3800csc(θ ). Thus
3800
ds = 3800 d csc(θ ) =−3800csc(θ )cot(θ )
dθ
dθ
π
When θ =
or 30° , then we have the triangle
3
π
3
2
π
6
1
3
Thus csc(θ ) = 2 and cot(θ ) = 3 so ds =−3800× 2 3 ≈−13164
dθ
Recall that the following two triangles give all the necessary
information for finding particular trigonometric functions.
π
4
2
1
π
4
1
π
3
2
π
6
3
1
Example. By calculating a few derivatives, and observing the
pattern, make a conjecture about
d 87 [sin(x)]
dx 87
d [sin( x)]= cos( x)
dx
d 2 [sin( x)]= d cos(x) =−sin(x)
dx 2
dx
d 3 [sin(x)] = d −sin(x) =−cos( x)
dx 3
dx
d 4 [sin(x)] = d −cos(x) = sin( x)
dx 4
dx
Then the pattern begins again and repeats forever. Thus
87
3







84







d [sin( x)] = d d [sin(x)] = d 3 [sin( x)] −cos(x)

dx87
dx3 dx84
dx3 