3.1 The Meaning of Slope of a Secant or a Tangent Line of a Function
The curve f has ”secant lines” L1; L2; L3 over the pairs of points (p0; p1); (p0; p2); (p0; p3) ; ; where pi = (xi; yi);
i = 0; 1; 2; 3; :::
The slope of these secant lines are xyi y0
i x0; i = 1; 2; 3; If the points p1; p2; p3; approaches p0 , and the slope xyi y0
i x0 has a limit T, then we say that the line passing though the point p0 = (x0; y0) of the curve f with the slope T is the ”tangent line”
of f at x0; y0). Figure.
Def. The slope of a tangent line of a function f at the point (c; f (c)) is de ned as
xlim!c
f (x) f (c)
x c = mT:
Note: The de nition above has an other form: Let x c = h:
Calculus 58 i.e. x = c + h: and x ! c is equivalent to h ! 0; i:e:
xlim!c
f (x) f (c)
x c = lim
h!0
f (c + h) f (c)
h :#
Ex. Figure.
If x denotes the displacement of an object, and t denotes the time after we measure this object, then the ”velocity” between time t1
and t2 is de ned as va = f (t2) f (t1)
t2 t1 :
It is the ”average velocity” from time t1 to time t2. If
tlim2!t1
f (t2) f (t1)
t2 t1 = vi exists, we called vi the ”instantaneous velocity” of the object at time t1.
Ex. In the previous example, if x denotes the velocity of an object, and t denotes the time, then the secant line's slope means the ”average acceleration” of this object. In case the tangent lines
exist at same points in f, the slope of a tangent line is called the
”instantaneous acceleration” of this object.
Note: In the above two examples, ”velocity” and ”acceleration”
are some kind of ”rate of change”. We usually regard ”rate of change” as ”instantaneous rate of change” rather than ”average
rate of change”.
Calculus 59 3.2 The Derivative
Def. The ”derivative” of a function f at a number c is de ned as Dxfjx=c = f0(c) =
Def. To nd a derivative, we usually call the process of nding it ”differentiation.”
Please read examples in the textbook.
Theorem: f0(c) exists) f : continuous at c.
Calculus 60 pf. To consider if f0(0) exists, simply think of lim
x!0+
jxj 0
x 0 and
xlim!0
jxj 0
x 0 , respectively.
* lim
Since left-hand limit 6= right-hand limit, by theorem we conclude that lim
x!0
jxj 0
x 0 doesn't exist,i.e. f0(0) doesn't exist; f cannot be differentiated at x = 0.#
Figure.
Calculus 61 3.3 Rules for Finding Derivatives
Theorem: f, g: differentiable (diff.) functions and k:constant, n 2 R n f0g. Then the following are true:
Calculus 62
Calculus 63
Calculus 64
*
( limh!0 g(x+h)1 = g(x)1 and
limh!0 g(x)g(x+h)f (x) = gf (x)2(x) both exist.
Thus we nd
hlim!0[(f
g)(x + h) (f
g)(x)]=h
= f0(x) g(x)
f (x)g0(x)
g2(x) = f0(x)g(x) f (x)g0(x) g2(x) :#
Derivatives of Sines and Cosines
? The sum of angles in sin and cos :
sin ( + ) = sin cos sin cos ; cos ( + ) = cos cos sin sin : Figure.
From the graph, we nd
sin tan ; 0 < 2;
and tan sin ; 2 0 (1):
From (1), sin (or sin ,)
Calculus 65 ) 0 sin 1 when 0 < j j < 2 (2)
Also from (1), tan = cossin (or tan ,) ) sin cos for 0 < j j < 2 (3).
Combining (2) and (3), we have 1 sin
cos (4) I. Now, consider the derivative of sin x : (sin x)0 = lim Let's check if lim
h!0
Hence by Squeeze Theorem, we nd
xlim!0
sin x
x = 1 from (4):
Calculus 66 iii) Finally, (5) becomes
hlim!0
Calculus 67
Calculus 68 3.4 The Chain Rule
Theorem: Let y = f (u) and u = g(x), F (x) = f g(x) = f (g(x)).
If g: differentiable at x & f: differentiable at u = g(x). )
F0(x) = (f g)0(x) = f0(g(x)) g0(x):
Proof: i) Intuitive thinking:
(f g)0(x) = lim
h!0
(f g)(x + h) (f g)(x) h
= lim
h!0[(f g)(x + h) (f g)(x) g(x + h) g(x)
g(x + h) g(x)
h ]:
* g : continuous,
) g(x + h) g(x) ! 0 as h ! 0. If g(x + h) 6= g(x); we have
(f g)0(x)
= lim
h!0
(f g)(x + h) (f g)(x)
g(x + h) g(x) lim
h!0
g(x + h) g(x) h
= f0(u)g0(x):#
However, we are not sure if
g(x + h) 6= g(x) or not.
Calculus 69 ii) Subtle proof: Let g(x) = u, and
q( ) = f (u + ) f (u)
f0(u); if 6= 0:
Then f (u + ) f (u) = [q( ) + f0(u)]. Note that lim!0q( ) = f0(u) f0(u) = 0;
hence we de ne q(0) = 0. (Therefore q( ) becomes continuous at 0.)
Now set = g(x + h) g(x), then
F (x + h) F (x) = f (u + ) f (u) = [q( ) + f0(u)];
and
F (x + h) F (x)
h =
h[q( ) + f0(u)];
) lim
h!0
F (x + h) F (x)
h = lim
h!0h[q( ) + f0(u)]
= lim
h!0
g(x + h) g(x)
h [q( ) + f0(u)]:
* g : continuous, i:e:
hlim!0g(x + h) = g(x);
and lim
h!0q( ) = lim
!0q( ) = 0:
Calculus 70
Calculus 71 ii)
(4x6 3x5 10x2 + 5x + 16)0
= (4x6)0 + ( 3x5)0 + ( 10x2)0 + (5x)0 + (16)0
= 24x5 15x4 20x + 5:#
Ex.
[(3x2 5)(2x4 x)]0 =?
Sol.
[(3x2 5)(2x4 x)]0
= (3x2 5)0(2x4 x) + (3x2 5)(2x4 x)0
= 6x(2x4 x) + (3x2 5)(8x3 1)
= 12x5 6x2 + 24x5 3x2 40x3 + 5
= 36x5 40x3 9x2 + 5:#
Note:
[(3x2 5)(2x4 x)]0 = (6x6 10x4 3x3 + 5x)0
= 36x5 40x3 9x2 + 5:#
Ex.
3x + 5 x2 + 7
0
=?
Calculus 72 Sol.
3x + 5 x2 + 7
0
= (3x + 5)0(x2 + 7) (3x + 5)(x2 + 7)0 (x2 + 7)2
= 3(x2 + 7) (3x + 5) 2x x4 + 14x2 + 49
= 3x2 + 21 6x2 10x x4 + 14x2 + 49
= 3x2 10x + 21 x4 + 14x2 + 49 :#
Ex.
(5x54)0 =?
Sol.
(5x54)0 = 5 5
4x54 1 = 25
4 x14:#
Ex.
(p5
x3)0 = (x35)0
= 3
5x35 1 = 3
5x 25:#
Ex.
(tan x2)0 =?
Sol.
(tan x2)0 = (sec2x2)2x:
Calculus 73 Reason: Let
tan x2 = f g(x);
where
f (u) = tan u; u = g(x) = x2: Hence by Chain Rule,
F (x) = tan x2 = f g(x);
) F0(x) = f0(u) g0(x)
= sec2(u) (2x) = (sec2x2) 2x:#
Ex.
(sin(cot x))0 =?
Sol. Let
sin(cot x) = f g(x);
where
f (u) = sin(u); u = g(x) = cot x:
) F(x) = sin cot x = f g(x);
F0(x) = f0(u)g0(x) = cos u ( csc2 x)
= cos(cot x) (csc2x):#
Ex.
(sec(cosp
t))0 =?
Calculus 74
Calculus 75 Then
F (x) = f g(x) = csc(5x2 + p3
5x);
F0(x) = f0(u) g0(x)
= csc(u) cot(u):(5x2 + p3 5x)0
= csc(5x2 + p3
5x) cot(5x2 + p3 5x) :(10x +
p3
5
3 x 23):#
Calculus 76 3.5 Leibniz Notation
Def. f (x) : a differentiable function w.r.t. x: Then f0(x)(= _f (x)) df
dx = d
dxf (x) :
This is called the ”Leibniz notation”. df is called the
”differential of f”, and similarly, dx is called the ”differential of x”.
Chain Rule:
dF
dx = df du
du , where dx
F (x) = f g (x) ; f = f (u) ; andu = g(x):
Calculus 77 3.6 Higher Order Differentiation
Def.
dnf
dxn f[n](x) = Dxnf (x)
= d dx( d
dx( df dx))
| {z }
n terms
= ((f0)
| {z }
n times
)0: Ex. Find
d2
dx2(5x2 + sin x):
Sol.
d2
dx2(5x2 + sin x) = (5x2 + sin x)"
= (10x + cos x)0 = 10 sin x: # Ex. Find
d3
dx3(tan x):
Calculus 78 Sol.
d3
dx3(tan x) = d2 dx2( d
dx tan x)
= d2
dx2(sec2 x) = d dx( d
dx sec2 x)
= d
dx(2 sec x sec x tan x)
= 4 sec x tan x sec x tan x + 2 sec2 x sec2x
= 2 sec2x(2 tan2x + sec2x): #
Calculus 79 3.7 Inverse Functions and Graph
Def. An inverse fun. f 1 of a function f : A ! B is de ned as f 1 : B ! A, and f 1 f = IA; f f 1 = IB:#
where IA(x) = x; 8x 2 A;
and IB(y) = y; 8y 2 B:
Theorem: For f 1 to exist, f must be 1-1.
pf. Assume that f 1 is a well-de ned relation, or (1) f 1(y1) 6= f 1(y2) ) y1 6= y2; 8yi 2 B Consider xi s.t. f (xi) = yi; i = 1; 2
Then (1) becomes
f 1(f (x1)) 6= f 1(f (x2))
) f(x1) 6= f(x2); 8xi 2 A , x1 6= x2 ) f(x1) 6= f(x2); 8xi 2 A , f is 1 1: #
Theorem : The graph of an inverse f 1 of f is G = f(f(x); x)jx 2 Rg
, where f : R ! R.
pf. A function f : R ! R has the graph G0 = f(x; f(x))jx 2 Rg
Calculus 80 , assume that f 1 exists. Then the graph of f 1 should be
G = f(y; f 1(y))jy 2 Rg:
Since y = f (x) for some x 2 R, f 1(y) = f 1(f (x)) = x:
) G = f(f(x); x)jx 2 Rg:
Ex. Figure.
Corollary: The Graph G of f 1 is symmetric to the graph G0 of f w.r.t. x = y in xy -plane.
pf. Suppose
G0 = f(x; y)jy = f(x)g;
then from the previous thm.
G = f(y; x)jy = f(x)g
The center of any point in G0 and a corresponding point in G is (x+y2 ; y+x2 ); y = f (x); i.e. The graph of the ”center” becomes
Gi = f(x + y
2 ; y + x
2 ))jy = f(x)g;
or
Gi = f(z; z)jz 2 Rg:
Calculus 81 i.e. x = y in xy-plane. #
Remark: From the idea of graph, we can have a more direct or geometric view of the inverse function. Moreover, even if f is not always 1-1, we may use graph to illustrate what f 1 might look like. Then we can determine where in the domain of f may f 1 exists. #
Calculus 82 3.8 Exponential Functions and Logarithmic Functions
Def. An exponential function is de ned as ax, where a is any constant, and x is a variable.
Discussion: if a 2 R, i) a = 0 :
) ax = 0x = 0 : constant function ii) a > 0 :
( ) 0 < a < 1 ( ) a = 1
( ) a > 1 Figure.
Remark: We usually consider the exponential function ax as a > 1. Since in case 0 < a < 1, we may consider a = a1
0, where a0 > 1. Also, only in the case a > 1 & 0 < a < 1 has ax an inverse function. As for the case that a < 0; it involves complex numbers and is out of our scoop of discussion here.
Def. When a > 1, we de ne the number e = a if
d
dxaxjx=0 = 1. Discussion:
Calculus 83
Def. The Logarithmic function loga x is de ned as the inverse function of ax; (a > 1), i.e.
loga(ax) = x; alogax = x:
Def. When a = e in the logarithmic function, we call the function
logex ln x the ”natural log function.”
Theorem:
i) loga x + loga y = loga xy ii) loga x loga y = loga xy iii). loga xk = k loga x; k 2 R:
Calculus 84 iv). loga b = loglogcb
ca: pf.
i) a(logax+logay)
= alogax alogay = x y
= a(logaxy):#
ii) a(logax logay)
= a(loga x)
a(loga y) = xy
= aloga xy:
iii). alogaxk = xk = alogax k = ak logax: # iv). Set x = loga b; then we nd that
logc b = logc (ax) = x logc a = loga b logc a;
) loga b = loglogcb
ca:
Hence i), ii), iii) and iv) all hold.
Calculus 85 3.9 Implicit Differentiation
For functions or equations that is not an explicit form of a speci c variable (or variables), it is dif cult to nd the differentiation w.r.t. this speci c variable(s). In these circumstances, we use
”implicit differentiation”.
e.g. 1) y3 + 7y = x3
It is dif cult (although not impossible) to nd y = y(x) and then compute y0 = dxdy.
2) y cot xy = 1
It is easy to show that
x = 1
y cot 1 1 y
yet it's dif cult to write y in terms of x, let along nd dydx.
Note: We will denote f (x)1 as (f (x)) 1 in the future, and f 1(x) as the inverse function of f (x).
Process of Implicit Differentiation:
1) Differentiate the whole eq. w.r.t. the considered variable ( ,say, x.)
2) Use Chain-Rule in differentiation: Consider the other form the ”implicit funs.” of x, and differentiate them.
3) Shift the terms with y0 together. Then we may nd y0 = f (x; y), for some f.
Ex. x2 + y2 = 25. Find y0(3).
Calculus 86 Sol. (x2 + y2)0 = (25)0 , 2x + 2yy0 = 0
i.e. y0 = xy.
Note that when x = 3, 32 + y2 = 25; ) y = 4. ) y0(3) = y(3)3 = 34:#
Ex. (xy) cot (xy) = 1. Find y0 =?
Sol. Use implicit differentiation : (xy cot (xy))0 = 10
, y cot (xy) + xy0 cot (xy) + xy( csc2 xy) (y + xy0) = 0
, y0(x cot (xy) x2y csc2 (xy))
= y cot (xy) + xy2 csc2(xy)
= y((csc2(xy)) xy cot (xy)) , y0 = y(xy csc2(xy) cot (xy))
x(cot (xy) xy csc2(xy))
= yx: #
Ex. x3 + y3 = 6xy. Find y0 in (3,3).
Sol. Figure.
* (x3 + y3)0 = (6xy)0
Calculus 87 , 3x2 + 3y2y0 = 6xy0 + 6y
, y0(6x 3y2) = 3x2 6y , y0(2x y2) = x2 2y ) y0 = x2x y2 2y2:
When x = 3 and y = 3 ) y0(3) = 9 66 9 = 1: #
Calculus 88 3.10 Differentiation of Inverse Functions.
3.10.1 Differentiation of inverse functions in general
Given a function f and its inverse function f 1 on I, if the inverse exists. Then the way to nd the derivative of f 1 is to combine the de nition of an inverse function as well as the implicit function differentiation method.
f f 1 (x) = x;
d
dxf f 1 (x) = df df 1
df 1
dx = 1:
We may de ne f 1 as a new variable u inside f; then the above relation becomes
df 1
dx = 1
df du
(|) ;
where f is considered as the function of u: We shall use this analogy in all sorts of differentiation involving inverse functions.
Ex. Given
y = xy tan x2 3y3 + x:
Note that y is a function of x: Assume that y 1(x) = u; please nd the differentiation of u; in terms of x and u:
Sol. First we have to nd the differentiation of f; namely y0 in our problem. However, y is considered as a function of u: i.e.
y = uy tan u2 3y3 + u:
Calculus 89 Use implicit function differentiation, we have
y0 = y tan u2 + uy0tan x2 + 2u2y sec2 u2 9y2y0 + 1;
) y0 = y tan u2 + 2u2y sec2u2 + 1 1 u tan u2 + 9y2 :
In order to nd the differentiation of y 1(x) = u; we will have to convert the equation into the form as in (|) : Now that the new variable is u; and
y (u) = y y 1 (x) = x;
we may replace x by u and y by x in (|) : i. e.
du
dx = 1 u tan u2 + 9x2
x tan u2 + 2u2x sec2u2 + 1: N
Calculus 90 3.10.2 Exponential functions and Logrithmic functions
1)
d
dxex = ex: (de ned.)
2)
d
dx(ln x) =? (1 x)
Discussion: Use inverse function and chain rule:
* eln x = x;
) (eln x) (ln x)0 = 1 = x(ln x)0 i. e.
(ln x)0 = 1 x: 3)
d
dxax =?
Sol.
ax = eln ax = ex ln a; ) d
dxax = d
dx(ex ln a) = ex ln a (ln a)
= (ln a) ex ln a = (ln a)ax:
Calculus 91
? Also from previous discussion, d
dxax = lim
h!0ax ah 1 h ; i.e.
hlim!0
ah 1
h = ln a: # Ex.
d
dx(ln(x5 + 3x + 1)) =?
Sol.
d
dx ln(x5 + 3x + 1)
= 1
x5 + 3x + 1 (x5 + 3x + 1)0
= 5x4 + 3
x5 + 3x + 1: # Simply set
ln(x5 + 3x + 1) = f g(x);
where
f (u) = ln u
&
g(x) = u = x5 + 3x + 1:
Calculus 92 Ex.
d
dx 23x2 =?
Sol. Let
23x2 = f g h(x);
where
f (u) = 2u; u = g(v) = 3v; v = x2 = h(x):
Then d
dx(23x2) = f0(u) g0(v) h0(x)
= (ln 2)2u (ln 3)3v (2x)
= (ln 2)(ln 3)23x2 3x2 (2x)
= 2(ln 2)(ln 3)x 3x2 23x2: # Ex.
d
dx [f (x)]g(x) =?
Sol. Note that
[f (x)]g(x) = exp n
ln [f (x)]g(x)o
= exp [g (x) ln f (x)] : So we have
d
dx [f (x)]g(x) = d
dx exp [g (x) ln f (x)]
Calculus 93
= exp [g (x) ln f (x)] g0 (x) ln f (x) + g0(x) f0 (x) f (x)
= [f (x)]g(x) g0 (x) ln f (x) + g0 (x) f0(x) f (x) :
Calculus 94 3.10.3 Inverse Trignometric Functions and Their
Derivatives 1)
sin 1 x; x 2 [ 1; 1]:
Figure.
2 Sin 1x
2; or
k 2 sin 1x k +
2; k 2 Z:
?: d
dx sin 1x =?
Sol. Let
y = sin 1 x: , sin y = x:
Use implicit differentiation:
(sin y)0 = (x)0 , y0 cos y = 1 , y0 = 1
cos y = 1
+p
1 sin2 y
Calculus 95
* k 1
2 sin 1x k + 1 2 for some xed k 2Z;
) y0 = 1
cos y =
( 1
p1 x2; k : even,
p 1
1 x2; k : odd.
i. e.
d
dx(sin 1x) = +1 p1 x2
+ : k even : k odd.
or
d
dx(Sin 1 x) = 1
p1 x2: # 2)
cos 1 x; x 2 [ 1; 1]:
Figure.
0 Cos 1x ; or
k cos 1 x k + ; k 2 Z:
Calculus 96
?: d
dx(cos 1x) =?
Sol. Let
y = cos 1x , cos y = x ) y0 = 1
sin y = +1
p1 cos2y
=
( +p 1
1 x2 ; k : odd
p 1
1 x2 ; k : even ; i. e.
d
dx(cos 1x) = 1 p1 x2
: k even + : k odd ; or
d dx(
1
Cos x) = 1
p1 x2: # 3)
tan 1x; 1 x 1:
Figure.
Calculus 97
2 tan 1x <
2; or
k 1
2 tan 1x < k + 1
2 ; k 2 Z:
?: d
dx tan 1x =?
Sol. Let
y = tan 1 x , tan y = x , y0 sec2y = 1 , y0 = 1
sec2 y ) y0 = 1
sec2y = 1
1 + tan2 y = 1 1 + x2: i. e.
d dx
1
Tan x = d
dx tan 1x = 1 1 + x2 4)
cot 1 x; 1 x 1:
Figure.
Calculus 98
0 Cot 1x ; or
k cot 1x (k + 1) :
?: d
dx(cot 1x) =?
Sol. Let
y = cot 1x , cot y = x
, y0 csc2 y = 1 , y0 = 1 csc2y: ) y0 = 1
csc2 y = 1
1 + cot2y = 1 1 + x2 i. e.
d
dx(Cot x) =1 d
dx cot 1x = 1
1 + x2: # 5)
sec 1 x; x 2 [ 1; 1] [ [1; 1]:
Figure.
Calculus 99
0 Sec 1 x ; or
k sec 1 x (k + 1) ; k 2 Z:
?: d
dx sec 1 x =?
Sol. Let
y = sec 1 x , sec y = x
, y0 sec y tan y = 1 , y0 = 1
sec y tan y: ) y0 = 1
sec y tan y = +1 xp
x2 1 (i) x > 0 (x 1) ) y 2 [0; 2] ) tan y > 0 i. e.
y0 = 1 xp
x2 1:
(ii) x < 0 (x 1) ) y 2 [2; ] ) tan y < 0 i. e.
y0 = 1
xp
x2 1:
Calculus 100 Hence we conclude that
d
dx(sec 1 x) = 1 jxjp
x2 1; or
d
dx(sec 1 x) =
( 1
jxjp
x2 1 ; k : even;
1 jxjp
x2 1 ; k : odd. # 6)
csc 1 x; x 2 [ 1; 1] [ [1; 1]:
Figure.
2 Csc 1 x
2; or
(k 1
2) csc 1x (k + 1
2) ; k 2 Z:
?: d
dx Csc 1x =?
Calculus 101 Hence we conclude that
d
Calculus 102 Sol.
d
dx(cot 12x) = (2x)0
1 + (2x)2 = (ln 2)2x
1 + 22x: # Ex.
d
dt(Sin 1(sec t)) =?
Sol.
d
dt(Sin 1(sec t)) = 1
p1 sec2 t(sec t)0
= sec t tan t p 1p
tan2t = i sec t: #
Calculus 103 3.11 Hyperbolic Functions & Inverse Hyperbolic Functions 3.11.1 Hyperbolic Functions
i) Hyperbolic Sine (sinh) sinh x = 1
2(ex e x); 8x 2 R:
ii) Hyperbolic Cosine (cosh) cosh x = 1
2(ex e x); 8x 2 R:
Figure.
iii) Hyperbolic Tangent (tanh) tanh x = sinh x
cosh x; 8x 2 R:
iv) Hyperbolic Cotangent (coth) coth x = cosh x
sinh x; 8x 2 Rnf0g:
Calculus 104 v) Hyperbolic Secant (sech)
sech x = 1
cosh x; 8x 2 R:
vi) Hyperbolic Cosecant (csch) csch x = 1
sinh x; 8x 2 Rnf0g:
Calculus 105 3.11.2 Special Relations among Hyperbolic Functions
i)
cosh2 x sinh2x = 1:
pf.
cosh2x sinh2x = (ex + e x
2 )2 (ex e x 2 )2
= 1
4(e2x + e 2x + 2 e2x e 2x ( 2)) = 1:
ii)
sech2x = 1 tanh2 x:
pf.
1 tanh2x = 1 (ex e x ex + e x)2
= 1 e2x + e 2x 2 e2x + e 2x + 2
= e2x + e 2x + 2 e2x e 2x + 2 e2x + e 2x + 2
= 4
e2x + e 2x + 2 = ( 2
ex + e x)2 = sech2 x: # iii)
csch2x = 1 + coth2x
Calculus 106 pf.
1 + coth2x = 1 + (ex + e x ex e x)2
= 1 + e2x + e 2x + 2 e2x + e 2x 2
= e2x + e 2x + 2 e2x e 2x + 2 e2x + e 2x 2
= 4
e2x + e 2x 2 = ( 2
ex + e x)2 = csch2x: #
Calculus 107 3.11.3 Differentiation for Hyperbolic Function
i) dxd (sinh x) = cosh x
pf. dxd (sinh x) = dxd (ex 2e x)
= 12(ex ( e x)) = 12(ex + e x) = cosh x: # ii) dxd (cosh x) = sinh x
pf. dxd (cosh x) = dxd (ex+e2 x)
= 12(ex + ( e x)) = 12(ex e x) = sinh x: # iii) dxd (tanh x) = dxd (cosh xsinh x)
= csch2x sinh2x
cosh2x = 1
cosh2 x = sech2x: # i.e. dxd (tanh x) = sech2x:
iv) dxd (coth x) = csch2 x
pf. dxd (cosh xsinh x) = sinh2x (cosh x)(sinh x)0 sinh2x
= 1
sinh2x = csch2x: # v) dxd (sech x) = (sech x)(tanh x) pf. dxd (sech x) = dxd [(cosh x) 1]
= (cosh x) 2 (sinh x) = cosh xsinh x cosh x1
= (sech x)(tanh x): #
vi) dxd (csch x) = (csch x)(coth x) pf. dxd (csch x) = dxd [(sinh x) 1]
= (sinh x) 2 (cosh x) = cosh xsinh x sinh x1
Calculus 108
= (csch x)(coth x): #
Calculus 109 3.11.4 Inverse Hyperbolic Functions
i) sinh 1x; 8x 2 R;
d
dx(sinh 1x) = 1
p1 + x2: pf. Let
y = sinh 1x , sinh y = x;
) 1 = y0 cosh y; y0 = 1 cosh y
= 1
p1 + sinh2 y = 1
p1 + x2: #
ii) cosh 1 x; cosh 1x 2 [ 1; 0] _ [0; 1]
8x 2 [1; 1]
d
dx(cosh 1 x) =
( 1
px2 1; if cosh 1 x 0;
p 1
x2 1; if cosh 1 x 0: :
Calculus 110 pf. Let
y = cosh 1x , cosh y = x , y0 sinh y = 1; y0 = 1
sinh y = +1
pcosh2y 1
=
( 1
px2 1; if cosh 1 x 0;
p 1
x2 1; if cosh 1 x 0: # iii) tanh 1x; tanh 1x 2 [ 1; 1]
8x 2 [ 1; 1];
d
dx(tanh 1x) = 1 1 x2: pf. Let
y = tanh 1x , tanh y = x , y0 sech2 y = 1
, y0 = 1
sech2 y = 1
1 tanh2y = 1
1 x2: # iv) coth 1 x; coth 1 x 2 [ 1; 1]
8x 2 [ 1; 1] [ [1; 1]
d
dx(coth 1 x) = 1 1 x2
Calculus 111
Calculus 112 pf. Let
y = csch 1x , csch y = x
, y0 csch y coth y = 1 , y0 = 1
csch y coth y
= 1
xp
x2 + 1
choose + if csch 1 x 0;
choose if csch 1 x 0: # Ex.
d
dx(atanh 1t3) =?
Sol. Let
F (t) = f g h(t);
where
f (u) = au; u = g(v) = tanh 1v;
v = h(t) = t3:
) F0(t) = f0(u) g0(v) h0(t)
= df du
dg dv
dh dt
= (ln a)(au) 1
1 v2 (3t2)
= (ln a) 3t2
1 t6 atanh 1t3: #
Calculus 113 Ex.
d
dx(sin 1 (tanh2 (sec x))) =?
Sol. Let
F (x) = f g h(x);
where
f (u) = sin 1 u; u = g(v) = tanh2v;
v = h(x) = sec x:
) F0(x) = df du
dg dv
dh dt
= +1
p1 u2 2 tanh v sech2v sec x tan x
= +2 tanh (sec x) sech2 (sec x) (sec x)(tan x)
(1 tanh4(sec x))12: #
Calculus 114