Download DIFFERENTIATION OF TRIGONOMETRIC FUNCTIONS

Differentiation of Trigonometric Functions
MODULE - V
Calculus
22
Notes
DIFFERENTIATION OF TRIGONOMETRIC
FUNCTIONS
Trigonometry is the branch of Mathematics that has made itself indispensable for other branches
of higher Mathematics may it be calculus, vectors, three dimensional geometry, functions-harmonic
and simple and otherwise just cannot be processed without encountering trigonometric functions.
Further within the specific limit, trigonometric functions give us the inverses as well.
The question now arises : Are all the rules of finding the derivatives studied by us so far appliacable
to trigonometric functions ?
This is what we propose to explore in this lesson and in the process, develop the formulae or
results for finding the derivatives of trigonometric functions and their inverses . In all discussions
involving the trignometric functions and their inverses, radian measure is used, unless otherwise
specifically mentioned.
.
OBJECTIVES
After studying this lesson, you will be able to :
l
l
l
l
find the derivative of trigonometric functions from first principle;
find the derivative of inverse trigonometric functions from first principle;
apply product, quotient and chain rule in finding derivatives of trigonometric and inverse
trigonometric functions; and
find second order derivative of a function.
EXPECTED BACKGROUND KNOWLEDGE
l
l
Knowledge of trigonometric ratios as functions of angles.
Standard limits of trigonometric functions namely.
sinx
tanx
= 1 (iii) limcosx = 1 (iv) lim
=1
x→ 0 x
x→ 0 x
x→ 0
(i) limsinx = 0 (ii) lim
x→ 0
l
Definition of derivative, and rules of finding derivatives of function.
MATHEMATICS
251
Differentiation of Trigonometric Functions
MODULE - V 22.1 DERIVATIVE OF TRIGONOMETRIC FUNCTIONS
Calculus
FROM FIRST PRINCIPLE
(i) Let y = sin x
For a small increment δx in x, let the corresponding increment in y be δy .
Notes
y + δy = sin(x + δx)
∴
δy = sin(x + δx) − sinx
and
δx 
δx

= 2cos  x +  sin
2
2

C+D
C −D

sinC − sinD = 2cos 2 sin 2 


δx
sin
δy
δ x

2
= 2cos  x + 
δx
2  δx

∴
δx
sin
δy
δx 

2
lim
= lim cos  x +
⋅ lim
2  δx→0 δx = cosx.1
δx →0 δx δx→ 0

2
Thus,
i.e.,
(ii) Let
sin δx




2
Q lim δx = 1
 δ x →0



2
dy
= cosx
dx
d
(sinx) = cosx
dx
y = cos x
For a small increment δx in x, let the corresponding increment in y be δy .
∴
and
y + δy = cos(x + δx)
δy = cos(x + δx) − cosx
δx 
δx

= −2sin  x +  sin
2
2

∴
δx
sin
δy
δx 

2
= −2sin  x +
⋅
δx
2  δx

δx
sin
δy
dx 

2
lim
= − lim sin  x +
⋅ lim

δ
x
2  δx→ 0
δx →0 δx
δ x→ 0

2
= − sinx ⋅1
Thus,
252
dy
= − sinx
dx
MATHEMATICS
Differentiation of Trigonometric Functions
(iii)
MODULE - V
Calculus
d
( cosx ) = − sinx
dx
i.e.,
Let
y = tan x
For a small increament δx in x, let the corresponding increament in y be δy .
y + δy = tan(x + x)
δ
∴
Notes
δy = tan(x + δx) −tanx
and
=
sin(x + δx) sinx
−
cos(x + δx) cosx
=
sin(x + δx) ⋅ cosx − sinx.cos(x + δx)
cos(x + δx)cosx
=
sin[(x + δx) − x ]
cos(x + δx)cosx
=
sin δx
cos(x + δx) ⋅ cosx
δy sin δ x
1
=
⋅
δx
δ x cos(x + δx)cosx
∴
δy
sin δ x
1
= lim
⋅ lim
δx→ 0 δx δ x→ 0 δ x δ x→ 0 cos(x + δx)cosx
lim
or
= 1⋅
1
2
cos x
sin δx 

= 1
Q δlim
 x →0 δx

= sec2 x
Thus,
dy
= sec2 x
dx
i.e.
d
(tanx) = sec2 x
dx
(iv) Let y = sec x
For a small increament δx in x, let the corresponding increament in y be δy .
∴
and
y + δy = sec(x + δx)
δy = sec(x + δx) − secx
1
1
= cos(x + δx) − cosx
=
cosx − cos(x + δx)
cos(x + δx)cosx
δx 
δx

2sin  x +  sin
2
2

=
cos(x + δ x)cosx
MATHEMATICS
253
Differentiation of Trigonometric Functions
MODULE - V
Calculus
δx

sin  x +  sin δx
δy
2 

2
lim
= lim
δx →0 δx δ x→ 0 cos(x + δx)cosx δx
2
δx 

δx
sin  x +
sin
δy
2 

2
lim
= lim
lim
δ
x
δx →0 δx δx→ 0 cos(x + δx)cosx δx →0
2
Notes
=
=
Thus,
i.e.
sinx
cos 2 x
⋅1
sinx
1
⋅
= tanx.secx
cosx cosx
dy
= secx.tanx
dx
d
(secx) = secx ⋅ tanx
dx
Similarly, we can show that
d
(cotx) = − cosec2 x
dx
and
d
(cosec x ) = − cosec x ⋅ cot x
dx
Example 22.1 Find the derivative of cotx2 from first principle.
Solution : y = cotx2
For a small increament δx in x, let the corresponding increament in y be δy .
∴
and
y + δy = cot(x + δx)2
δy = cot(x + δx) 2 − c o t x2
=
=
=
=
254
cos(x + δx)2
sin(x + δx)2
−
cosx2
sinx2
cos(x + δx) 2 sinx 2 − cosx 2 sin(x + δx) 2
sin(x + δx)2 sinx2
sin[x 2 − (x + δx)2 ]
sin(x + δx)2 sinx2
sin[−2xδx − (δ x) 2 ]
sin(x + δx)2 sinx 2
MATHEMATICS
Differentiation of Trigonometric Functions
=
MODULE - V
Calculus
− sin[(2x + δ x)δx]
sin(x + δx)2 sinx2
δy
− sin[(2x + δx)δx]
=
δx δxsin(x + δx )2 sinx2
∴
δy
sin[(2x + δx) δx]
2x + δx
= − lim
lim
δx →0 δx
δ x→ 0 δ x(2x + δx) δ x→ 0 sin(x + δx)2 sinx2
lim
and
2x
dy
= −1 ⋅
2
dx
sinx .sinx2
or
=
−2x
2 2
(sinx )
=
Notes

sin[(2x + δx) δx] 
= 1
Q δlim
 x →0 δx(2x + δx)

−2x
sin 2 x 2
= −2x.cosec2 x 2
d
(cotx 2 ) = −2x ⋅ cosec2 x 2
dx
Hence
Example 22.2 Find the derivative of
cosecx from first principle.
Solution : Let y = cosecx
y + δ y = cosec(x + δx)
and
 cosec(x + δx) − cosecx   cosec(x + δx ) + cosec x 
δy = 
cosec(x + δx) + cosecx
∴
=
cosec(x + δx) − cosecx
cosec(x + δx) + cosecx
=
1
1
−
sin(x + δx) s i n x
cosec(x + δx) + cosecx
=
sinx − sin(x + δx)
 cosec(x + δx) + cosecx [ sin ( x + xδ) sinx]
δx
δx
2cos  x +  sin

2
2
= −
[
cosec(x
+
δ
x)
+
cosecx
sin
(
) ( x + xδ)sinx]
δy
= − lim
δx →0 δx
δx→ 0
lim
or
MATHEMATICS
δx 

sin δx / 2
cos  x + 
2 

δx / 2
×
cosec(x + δ x) + cosecx] [sin(x + δx).sinx]
dy
− cosx
=
dx (2 (cosecx)(sinx)2
255
Differentiation of Trigonometric Functions
MODULE - V
Calculus
1
−
1
= − (cosecx) 2 (cosec x c o t x )
2
Thus,
Notes
d
dx
(
)
1
1
−
cosecx = − ( cosecx ) 2 ( cosecxcotx)
2
Example 22.3 Find the derivative of sec2 x from first principle.
Solution : Let y = sec 2 x
y + δy = sec 2 (x + δx)
and
δy = sec 2 (x + δ x) − sec 2 x
then,
=
=
=
cos 2 x − cos2 (x + δx)
cos 2 (x + δx)cos 2 x
sin[(x + δ x + x]sin[(x + δx − x)]
cos2 (x + δx)cos2 x
sin(2x + δx)sin δx
cos 2 (x + δx)cos 2 x
δy
sin(2x + δx)sin δx
=
δx cos2 (x + δx)cos2 x.δx
lim
Now,
δy
δx →0 δx
= lim
sin(2x + δx)sin δx
δx→0 cos 2 (x + δx)cos2
x.δx
dy
sin2x
=
2
dx cos xcos2 x
=
2sinxcosx
2
2
cos xcos x
= 2tanx.sec 2 x
= 2secx(sec x.tanx)
= 2sec x (sec x tan x)
CHECK YOUR PROGRESS 22.1
1.
2.
Find the derivative from first principle of the following functions with respect to x :
(a) cosec x
(b) cot x
(c) cos 2 x
(d) cot 2 x
(e) c o s e c x2
(f) s i n x
Find the derivative of each of the following functions :
(a) 2sin 2 x
256
(b) cosec2 x
(c) tan 2 x
MATHEMATICS
Differentiation of Trigonometric Functions
MODULE - V
Calculus
22.2 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
You have learnt how we can find the derivative of a trigonometric function from first principle
and also how to deal with these functions as a function of a function as shown in the alternative
method. Now we consider some more examples of these derivatives.
Example 22.4 Find the derivative of each of the following functions :
(i) sin 2 x
(ii) tan x
Solution : (i) Let
(iii) cosec(5x 3 )
y = sin 2 x,
= sin t,
where t = 2 x
dy
= cost
dt
By chain Rule,
Notes
and
dt
=2
dx
dy dy dt
= ⋅ , we have
dx dt dx
dy
= cos t (2) = 2. cos t = 2 cos 2 x
dx
Hence,
d
(sin2x) = 2cos2x
dx
(ii) Let
y = tan x
= tan t
where t = x
dy
= sec2 t and
dt
∴
dt
1
=
dx 2 x
dy dy dt
= ⋅ , we have
dx dt dx
By chain rule,
1
dy
sec2 x
= sec 2 t ⋅
=
2 x
dx
2 x
d (
sec 2 x
tan x ) =
dx
2 x
Hence,
Alternatively : Let y = tan x
dy
d
sec2 x
= sec2 x
x =
dx
dx
2 x
(iii) Let y = cosec(5x 3 )
dy
d
= − cosec(5x 3 )cot(5x 3 ) ⋅ [5x 3 ]
dx
dx
∴
= −15x 2 cosec(5x 3 )cot(5x 3 )
or
you may solve it by substituting t = 5x 3
MATHEMATICS
257
Differentiation of Trigonometric Functions
MODULE - V
Calculus
Example 22.5
Find the derivative of each of the following functions :
sinx
(ii) y = 1 + cosx
(i) y = x 4 sin2x
Solution :
Notes
(i)
∴
y = x 4 sin2x
dy
d
d
= x 4 (sin2x) + sin2x (x 4 )
dx
dx
dx
(Using product rule)
= x 4 (2cos2x) + sin2x(4x 3 )
= 2x 4 cos2x + 4x 3 sin2x
= 2x 3[xcos2x + 2sin2x]
(ii)
y=
=
sinx
1 + cosx
x
x
cos
2
2
2x
2cos
2
2sin
= tan
∴
x
2
dy
x d  x 1
x
= sec 2 ⋅   = sec2
dx
2 dx  2  2
2
Alternatively : You may find the derivative by using quotient rule
Let
y=
∴
dy
=
dx
sin x
1 + cos x
(1 + cosx)
=
=
=
=
=
258
d
d
(sinx) − sin x (1 + cos x )
dx
dx
(1 + cos x ) 2
(1 + cosx)(cosx) − sin x ( − sin x )
(1 + cos x ) 2
cosx + cos 2 x + sin 2 x
(1 + cos x)2
cosx + 1
(1 + cos x ) 2
1
(1 + cos x )
1
x
2cos 2
2
1
x
= sec2
2
2
MATHEMATICS
Differentiation of Trigonometric Functions
Example 22.6
Find the derivative of each of the following functions w.r.t. x :
(i) cos2 x
MODULE - V
Calculus
(ii) sin 3 x
Solution : (i) Let y = cos 2 x
= t2
where t = cos x
dy
= 2t
dt
∴
Notes
dt
= − sin x
dx
and
Using chain rule
dy dy dt
= ⋅ , we have
dx dt dx
dy
= 2 cos x. ( − sinx)
dx
= −2cosxsinx = − sin2x
y = sin 3 x
(ii) Let
dy 1
d
= (sin 3 x)−1 / 2 ⋅ (sin 3 x)
dx 2
dx
∴
1
=
3
⋅ 3sin 2 x ⋅ cosx
2 sin x
=
3
sin x cos x
2
d 
3  3
sin x cos x
 sin x  =
 2
dx 
Thus,
Example 22.7 Find
y=
(i)
dy
, when
dx
1 − sin x
1 + sin x
(ii)
y = a(1 − cost),x = a(t + sint)
Solution : We have,
(i)
∴
y=
1 − sin x
1 + sin x
dy 1  1 − sin
=
dx 2  1+ sin
MATHEMATICS
x
x 
−
1
2
⋅
d 1 − sinx 
dx  1+ sinx 
259
Differentiation of Trigonometric Functions
MODULE - V
Calculus
Notes
=
1 1 + sin x ( − cosx)(1 + sinx) − (1 − sinx)(cosx)
⋅
2 1 − sin x
(1 + sinx)2
=
1 1 + sin x  −2cosx 
⋅

2 1 − sin x  (1 + sinx)2 
=−
=−
Thus, dy/dx = −
1 + sinx 1 − sin 2 x
⋅
1 − sinx (1 + sinx) 2
1 + sinx 1 + sinx
(1 + sinx)
2
=
−1
1 + sinx
1
1 + sinx
Alternatively, it is more convenient to find the derivative of such square root function by
rationalising the denominator.
1 − sinx
1 − sinx
×
1 + sinx
1 − sinx
y=
=
=
1 − sinx
1 − sin 2 x
1 − sinx
cos x
= secx − tanx
∴
dy
sin x
1
= sec x t a n x − sec2 x =
−
2
dx
cos x cos2 x
=
(ii)
∴
sin x − 1
1 − sin x
2
=−
1
1 + sin x
x = a(t + sint), y = a(1 − cost)
dx
= a(1 + cost),
dt
Using chain rule,
dy
= a(sint)
dt
dy dy dt
= ⋅
, we have
dx dt dx
dy
a(sint)
=
dx a(1 + cost)
=
260
t
t
cos
2
2
t
= tan
t
2
2cos
2
2
2sin
MATHEMATICS
Differentiation of Trigonometric Functions
Find the derivative of each of the following functions at the indicated points : MODULE - V
Example 22.8
(i)
y = sin2x + (2x − 5)
(ii)
y = cotx + sec 2 x + 5
2
Calculus
π
at x =
2
at x = π / 6
Solution :
Notes
y = sin2x + (2x − 5) 2
(i)
dy
d
d
= cos2x
(2x) + 2(2x − 5) (2x − 5)
dx
dx
dx
= 2cos2x + 4(2x − 5)
∴
At x =
π
,
2
dy
= 2cos π + 4( π − 5)
dx
= −2 + 4π − 20
= 4π − 22
y = cotx + sec 2 x + 5
(ii)
dy
= − cosec 2 x + 2secx(secxtanx)
dx
∴
= − cosec 2 x + 2sec2 xtanx
At x =
π
,
6
dy
π
π
π
= − cosec 2 + 2sec 2 tan
dx
6
6
6
= −4 + 2 ⋅
4 1
3 3
8
= −4 +
3 3
Example 22.9 If sin y = x sin (a+y), prove that
2
dy sin (a + y )
=
dx
sina
Solution : It is given that
sin y = x sin (a+y)
or
x=
siny
sin(a + y)
.....(1)
Differentiating w.r.t. x on both sides of (1) we get
 sin(a + y)cosy − sinycos(a + y)  dy
1= 

sin 2 (a + y)

 dx
or
MATHEMATICS
 sin(a + y −y)  dy
1= 

2
 sin (a + y)  dx
261
Differentiation of Trigonometric Functions
2
dy sin (a + y )
=
dx
sina
MODULE - V
or
Calculus
Example 22.10 If y = sinx + sinx + ....to infinity ,
dy cosx
=
dx 2y − 1
prove that
Notes
Solution : We are given that
y = sinx + sinx + ...toinfinity
y = sin x + y
or
y 2 = sinx + y
or
Differentiating with respect to x , we get
2y
dy
dy
= cosx +
dx
dx
(2y − 1)
or
dy
= cosx
dx
dy cosx
=
dx 2y − 1
Thus,
CHECK YOUR PROGRESS 22.2
1.
2.
Find the derivative of each of the following functions w.r.t x :
(a) y = 3sin4x
(b) y = cos5x
(c) y = tan x
(d) y = sin x
(e) y = sinx2
(f) y = 2 tan2x
(g) y = π cot3x
(h) y = sec10x
(i) y = cosec2x
Find the derivative of each of the following functions :
(a) f(x) =
secx − 1
secx + 1
(
(b) f(x) =
sinx + cosx
sinx − cosx
)
2
(d) f(x) = 1 + x cosx (e) f(x) = x cosecx
(c) f(x) = x sinx
(f) f(x) = sin2xcos3x
(g) f(x) = sin3x
3.
262
Find the derivative of each of the following functions :
(a)
y = sin3 x
(b)
y = cos 2 x
(c)
y = tan 4 x
(d)
y = cot 4 x
(e)
y = sec5 x
(f)
y = cosec 3 x
(g)
y = sec x
(h)
y=
secx + tanx
secx − tanx
MATHEMATICS
Differentiation of Trigonometric Functions
4.
(a)
5.
MODULE - V
Calculus
Find the derivative of the following functions at the indicated points :
π
y = cos(2x + π/ 2),x =
3
y=
(b)
1+ s i n x
π
,x =
cosx
4
If y = tanx + tanx + tanx + ..., to infinity
dy
= sec 2 x .
dx
If cosy = xcos(a + y),
Notes
Show that (2y − 1)
6.
prove that
dy cos2 (a + y)
=
.
dx
sina
22.3 DERIVATIVES OF INVERSE TRIGONOMETRIC
FUNCTIONS FROM FIRST PRINCIPLE
We now find derivatives of standard inverse trignometric functions sin −1 x, cos− 1 x, tan −1 x, by
first principle.
(i) We will show that by first principle the derivative of sin −1 x w.r.t. x is given by
d
(sin − 1 x) =
dx
1
(1 − x 2 )
Let y = sin −1 x . Then x = sin y and so x + δx = sin(y + δy)
As δx → 0, δy → 0 .
Now,
δx = sin(y + δy) − sin y
∴
1=
sin(y + δy) − s i n y
δx
or
1=
sin(y + δy) − sin y δy
⋅
δy
δx
∴
1 = lim
[On dividing both sides by δx ]
sin(y + δy) − siny
δy
⋅ lim
δ y→0
δy
δx →0 δx
[Q δ y → 0 when δx → 0]

1 

 1 
2cos  y + δ y  sin  δ y  

2 

 2   ⋅ dy
=  lim
 δ y→ 0
δy
 dx
dy
= ( cosy) ⋅
dx
dy
1
1
=
=
dx cos y
1 − sin 2 y
(
MATHEMATICS
=
1
) (1 − x 2 )
263
Differentiation of Trigonometric Functions
MODULE - V
Calculus
∴
d ( −1 )
sin x =
dx
(1 − x 2 )
(
)
d
1
cos −1 x = −
dx
1 − x2
(ii)
Notes
1
(
)
⋅
For proof proceed exactly as in the case of sin −1 x .
(iii)
Now we show that,
(
)
d
1
tan − 1 x =
dx
1 + x2
Let y = tan −1 x .Then x = tany and so x + δx = tan(y + δy)
As δx → 0, also δy → 0
Now,
δx = tan(y + δy) − t a n y
tan(y + δy) − tany δy
⋅ ⋅
δy
δx
∴
1=
∴
1 = lim
tan(y + δy) − tany
δy
⋅ lim
δ y→ 0
δy
δx →0 δx
[Qδ y → 0 when δx → 0 ]

 dy
 sin(y + δy) siny 
=  lim 
−
 δy  ⋅
δy→0  cos(y + δy) cosy
 dx
(iv)
dy
sin(y + δ y)cosy − cos(y + δy)siny
⋅ lim
dx δy →0
δy.cos(y + δy)cosy
=
dy
sin(y + δy − y)
⋅ lim
dx δy →0 δy.cos(y + δy)cosy
=
dy
 sin δy
1

⋅ lim 
⋅
dx δy →0  δy cos(y + δy)cosy
=
dy
1
dy
⋅ 2 =
⋅ sec 2 y
dx cos y dx
dy
1
1
1
=
=
=
2
2
dx sec y 1 + tan y 1 +x 2
∴
∴
=
⋅
d ( −1 )
1
tan x =
dx
1+ x2
(
)
d
1
cot − 1 x = −
dx
1 + x2
For proof proceed exactly as in the case of tan −1 x .
264
MATHEMATICS
Differentiation of Trigonometric Functions
(v)
d
(sec− 1 x) =
We have by first principle
dx
x
1
(x 2 − 1 )
MODULE - V
Calculus
Let y = sec − 1 x . Then x = sec y and so x + δx = sec(y + δy).
As δx → 0,also δy → 0 .
Notes
δx = sec(y + δy ) − secy.
Now
sec(y + δy) − secy δy
⋅ ⋅
δy
δx
1=
∴
sec(y + δ y) − secy
δy
⋅ lim
δy
δx →0 δx
1 = lim
δ y→ 0
dy
=
⋅ lim
dx δy →0
[Qδ y → 0 when δx → 0 ]
1  1 

2sin  y + δy  sin  δy 
2  2 

δy.cosycos ( y + δy )

1
1 
sin  y + δy
sin
δy

dy
2 
2  



=
⋅ lim 
⋅
1
dx δ y→0  cosycos ( y + δy )
δy 


2
dy
siny
dy
= dx ⋅ cosycosy = dx ⋅ secytany
dy
1
=
=
dx secytany secy
∴
(sec2 y − 1)
1
=
x
( x2 − 1)
d ( −1 )
1
= sec x =
dx
x x 2 −1
∴
(vi)
1
(
)
d
cosec − 1x =
dx
x
1
( x − 1)
.
2
For proof proceed as in the case of sec−1 x.
Example 22.11 Find derivative of sin − 1 ( x 2 ) from first principle.
Solution :
y = sin −1 x 2
Let
x2 = s i n y
∴
( x + δx )2 = sin(y + δy)
Now,
( x + δx )2 − x 2
δx
MATHEMATICS
=
sin ( y + δy ) − siny
δx
265
Differentiation of Trigonometric Functions
MODULE - V
Calculus
lim
δx →0
δy 

2cos  y +  sin δy
2

2 ⋅ lim δy
= lim
δy δx→0 δx
(x + δx) − x
δy ←0
2
2
( x + δx )2 − x 2
dy
dx
⇒
2x = cosy⋅
⇒
dy
2x
2x
2x
=
=
=
.
dx cosy
1 − sin 2 y
1− x 4
Notes
Example 22.12 Find derivative of sin −1 x w.r.t. x by delta method.
Solution : Let y = sin −1 x
⇒
siny = x
..(1)
Also
sin(y + δy) = x + δx
From (1) and (2), we get
..(2)
sin(y + δy) − siny = x + δx − x
or
δy   δy 

2cos  y +  sin   =
2  2

=
∴
δ y  δ y

2cos  y +  sin  

2 
 2 =
δx
x + δ x− x
δx
x + δ x+
∴
∴
1
x + δx + x
dy
1
cosy=
or
dx
2 x
)
1
x + δx + x
δy
δy 

lim
⋅ lim cos  y +  ⋅ lim
δ x →0 δ x δ y →0
2  δ y→ 0

or
x + δ x+ x
x
or
δ x →0
)(
x + δx + x
 δy 
sin  
δy
δy 

 2=
⋅ cos  y +
 ⋅ δy
δx
2 

2
= lim
266
(
1
x + δx + x
 δy 
sin  
 2 
δy
2
(Q
δ y → 0 as δ x → 0)
dy
1
1
1
=
=
=
dx 2 x cosy 2 x 1− sin 2 y 2 x 1− x
dy
1
=
dx 2 x 1 − x
MATHEMATICS
Differentiation of Trigonometric Functions
MODULE - V
Calculus
CHECK YOUR PROGRESS 22.3
1. Find by first principle that derivative of each of the following :
(i) cos−1 x2
(ii)
cos− 1 x
x
−1
(iv) tan −1 x 2
(v)
tan x
x
(iii) cos−1 x
Notes
(vi) tan −1 x
22.4 DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS
In the previous section, we have learnt to find derivatives of inverse trignometric functions by
first principle. Now we learn to find derivatives of inverse trigonometric functions by alternative
methods. We start with standard inverse trignometric functions sin − 1 x,cos−1 x,....
−1
(i) Derivative of sin x
Solution : Let y = sin −1 x
x = sin y
∴
Differentiating w.r.t. y
(i)
dx
= cosy
dy
∴
dx
= 1 − sin 2 y
dy
or
1
1
=
dx
1 − sin 2 y
dy
∴
dy
1
1
=
=
2
dx
1 − sin y
1 − x2
Hence,
...[Using (i) ]
 dy 1 
Q dx = dx 


dy 

d
1
[sin − 1 x] =
dx
1− x2
Similarly we can show that
d
−1
[cos−1 x] =
dx
1 − x2
(ii) Derivative of tan −1 x
Solution : Let tan − 1 x = y
∴
x = tany
dx
2
Differentiating w.r.t. y, dy = sec y
MATHEMATICS
267
Differentiation of Trigonometric Functions
MODULE - V
and
Calculus
dy
1
=
dx sec 2 y
=
=
Notes
(
)
(
)
1
[Q We have written tan y in terms of x]
1 + tan 2 y
1
1+ x2
Hence,
d
1
tan − 1 x =
dx
1+ x2
Similarly,
d
−1
cot −1 x =
.
dx
1 + x2
(iii) Derivative of sec−1 x
Solution : Let sec −1 x = y
dx
= secytany
dy
∴
x = sec y and
∴
dy
1
=
dx sec y tan y
=
=
=
1
sec y [ ± sec 2 y − 1]
1
± sec y sec 2 y − 1
1
| x | sec 2 y − 1


π
2
Note : (i) When x >1, sec y is + ve and tan y is + ve, y ∈  0, 
π
2
 
(ii) When x < −1, sec y is −ve and tan y is −ve, y ∈  , π
Hence,
d
1
(sec− 1 x) =
dx
| x | x2 − 1
Similarly
d
−1
(cosec−1x) =
dx
| x | x2 − 1

Example 22.13 Find the derivative of each of the following :
(i) sin −1 x
(ii) cos−1 x2
(iii) (cosec − 1x)2
Solution :
(i)
268
Let
y = sin −1 x
MATHEMATICS
Differentiation of Trigonometric Functions
dy
=
dx
∴
1
1−
( )
x
2
d
dx
MODULE - V
Calculus
( )
x
1
1
⋅ x −1 / 2
1− x 2
=
Notes
1
=
2 x 1− x
d
1
sin − 1 x =
dx
2 x 1− x
∴
(ii) Let y = cos −1 x 2
dy
−1
d
=
⋅ (x 2 )
dx
1 − (x 2 )2 dx
=
−1
1 − x4
⋅ (2x)
d ( −1 2 )
−2x
cos x =
dx
1 − x4
∴
(iii) Let y = (cosec −1x) 2
(
dy
d
= 2(cosec−1x) ⋅
cosec− 1x
dx
dx
= 2(cosec− 1x) ⋅
=
)
−1
| x | x 2 −1
−2cosec − 1x
| x | x2 − 1
d
−2cosec − 1x
(cosec− 1x)2 =
dx
| x | x2 − 1
∴
Example 22.14 Find the derivative of each of the following :
(i)
tan −1
cosx
1 + sinx
(ii) sin(2sin −1 x)
Solution :
Let (i)
y = tan−1
cosx
1 + sinx
π 
sin  − x
2

= tan
π 
1 + cos  −x
2

−1
MATHEMATICS
269
Differentiation of Trigonometric Functions
MODULE - V
Calculus
  π x 
= tan −1 tan  −  
  4 2 
=
∴
Notes
π x
−
4 2
dy
= −1/2
dx
(ii)
y = sin(2sin −1 x)
Let
y = sin(2sin −1 x)
∴
dy
d
= cos(2sin −1 x) ⋅ (2sin −1x)
dx
dx
∴
dy
2
= cos(2sin −1 x) ⋅
dx
1 − x2
=
2cos(2sin −1 x)
1− x2
−1
Example 22.15 Show that the derivative of tan
Solution : Let
y = tan −1
2x
1− x
Let
x = tan θ
∴
y = tan−1
2
and z = sin−1
2tan θ
2
1 − tan θ
2x
1− x
2
w.r.t sin −1
2x
1 + x2
is 1.
2x
1 + x2
and z = sin −1
2tan θ
1 + tan 2 θ
= tan −1 (tan2θ) and z = sin − 1(sin2 θ)
= 2θ
and
dy
=2
dθ
and
z = 2θ
dz
=2
dθ
dy dy d θ
1
= ⋅
= 2 ⋅ =1
dx dθ dz
2
(By chain rule)
CHECK YOUR PROGRESS 22.4
Find the derivative of each of the following functions w.r.t. x and express the result
in the simplest form (1-3) :
270
x
2
1.
(a) sin −1 x 2
(b) cos −1
(c) cos −1
2.
(a) tan −1 (cosecx − cotx)
(b) cot −1 (secx + tanx) (c) tan
1
x
−1 cosx − sinx
cosx + sinx
MATHEMATICS
Differentiation of Trigonometric Functions
3.
4.
(a) sin(cos− 1 x)
(b) sec(tan − 1 x)
(d) cos− 1(4x 3 − 3x)
(e) cot −1  1 + x 2 + x 

MODULE - V
Calculus
(c) sin − 1(1 − 2x 2 )

Find the derivative of :
tan −1 x
1 + tan
−1
x
w.r.t. tan −1 x .
Notes
22.5 SECOND ORDER DERIVATIVES
We know that the second order derivative of a function is the derivative of the first derivative of
that function. In this section, we shall find the second order derivatives of trigonometric and
inverse trigonometric functions. In the process, we shall be using product rule, quotient rule and
chain rule.
Let us take some examples.
Example 22.16 Find the second order derivative of
(i) s i n x (ii) xcosx
(iii) cos−1 x
Solution : (i) Let y = sin x
Differentiating w.r.t. x both sides, we get
dy
= cosx
dx
Differentiating w.r.t x both sides again, we get
d2 y
dx
∴
2
d2 y
dx 2
=
d
(cosx) = − sinx
dx
= − sinx
(ii) Let y = x cos x
Differentiating w.r.t. x both sides, we get
dy
= x( − sinx) + cosx.1
dx
dy
= − xsinx + cosx
dx
Differentiating w.r.t. x both sides again, we get
d2 y
dx
2
=
d
( − xsinx + cosx )
dx
= − ( x.cosx + sinx ) − sinx
= − x.cosx − 2sinx
∴
MATHEMATICS
d2 y
dx 2
= − ( x.cosx + 2sinx)
271
Differentiation of Trigonometric Functions
MODULE - V (iii) Let y = cos−1 x
Calculus
Differentiating w.r.t. x both sides, we get
(
dy
−1
−1
=
=
= − 1 − x2
1
/
2
dx
1 − x2
1 − x2
(
Notes
)
)
−
1
2
Differentiating w.r.t. x both sides, we get
(
d2 y
 −1
= −  ⋅ 1 − x2
2
dx
2
=−
d2 y
∴
dx
2
=
)
−3 / 2

⋅ ( −2x ) 

x
(1 − x 2 )
3/2
−x
(1 − x2 )
3/2
(
)
2
Example 22.17 If y = sin −1 x , show that 1 − x y 2 − xy1 = 0 , where y 2 and y1 respectively
denote the second and first, order derivatives of y w.r.t. x.
Solution : We have,
y = sin −1 x
Differentiating w.r.t. x both sides, we get
dy
1
=
dx
1 − x2
2
or
or
1
 dy 
 dx  =
  1 − x2
(squaring both sides)
(1 − x2 ) y12 = 0
Differentiating w.r.t. x both sides, we get
or
or
(1 − x 2 ) ⋅ 2y1 dxd ( y1 ) + ( −2x )⋅ y12 = 0
(1 − x 2 ) ⋅ 2y1y2 − 2 x y12 = 0
(1 − x2 ) y2 − x y1 = 0
CHECK YOUR PROGRESS 22.5
1.
272
Find the second order derivative of each of the following :
(a) sin(cosx)
(b) x 2 tan −1 x
MATHEMATICS
Differentiation of Trigonometric Functions
2.
If y =
(
1
sin − 1 x
2
)
2
(
MODULE - V
Calculus
)
2
, show that 1 − x y2 − xy1 = 1 .
d2 y
+ tanx
3.
If y = sin(sinx) , prove that
4.
2
If y = x + tanx, show that cos x
dx
2
d2 y
dx 2
dy
+ ycos2 x = 0 .
dx
− 2y + 2x = 0
Notes
LET US SUM UP
●
●
●
●
(i)
d
(sinx) = cosx
dx
(ii)
d
(cosx) = − sinx
dx
(iii)
d
(tanx) = sec2 x
dx
(iv)
d
(cotx) = − cosec2 x
dx
(v)
d
(secx) = secxtanx
dx
(vi)
d
(cosecx) = − cosecxcotx
dx
If u is a derivabale function of x, then
(i)
d
du
(sinu) = cosu
dx
dx
(ii)
d
du
(cosu) = − sinu
dx
dx
(iii)
d
du
(tanu) = sec2 u
dx
dx
(iv)
d
du
(cotu) = − cosec2 u
dx
dx
(v)
d
du
(secu) = secutanu
dx
dx
(vi)
d
du
(coseu) = −cosec u c o t u
dx
dx
(i)
d
1
(sin − 1 x) =
dx
1 − x2
(ii)
d
−1
(cos− 1 x) =
dx
1 − x2
(iii)
d
1
(tan −1 x) =
dx
1 + x2
(iv)
d
−1
(cot −1 x) =
dx
1 + x2
(v)
d
1
(sec− 1 x) =
dx
| x | x2 − 1
(vi)
d
−1
(cosec−1x) =
dx
| x | x2 − 1
If u is a derivable function of x, then
(i)
d
1
du
(sin − 1 u) =
⋅
2
dx
1 − u dx
(ii)
d
−1
du
(cos− 1 u) =
⋅
2
dx
1 − u dx
(iii)
d
1
du
(tan −1 u) =
⋅
2
dx
1 + u dx
(iv)
d
−1 du
(cot −1 u) =
⋅
dx
1 + u 2 dx
(v)
d
1
du
(sec− 1 u) =
⋅
(vi)
dx
| u | u 2 − 1 dx
d
−1
du
(cosec− 1u) =
⋅
dx
| u | u 2 − 1 dx
The second order derivative of a trignometric function is the derivative of their first order
derivatives.
MATHEMATICS
273
Differentiation of Trigonometric Functions
MODULE - V
Calculus
SUPPORTIVE WEB SITES
l
l
http://www.wikipedia.org
http://mathworld.wolfram.com
Notes
TERMINAL EXERCISE
x
dy
, find
.
2
dx
1.
If y = x 3 tan2
2.
Evaluate,
3.
If y =
4.
If y = sec −1
5.

If x = acos3 θ, y = asin 3 θ , then find 1 + 
6.
If y = x + x + x + .... , find
.
dx
7.
Find the derivative of sin −1 x w.r.t. cos− 1 1 − x 2
8.
If y = cos(cosx) , prove that
d
π
sin 4 x + cos 4 x at x = and 0.
dx
2
5x
3
(1 − x)
+ cos2 (2x + 1) , find
dy
.
dx
dy
x +1
x −1
+ sin −1
, then show that
=0
dx
x −1
x +1
2
dy 
 .
 dx 
dy
d2 y
dx
9.
2
2
− cotx ⋅
dy
+ y.sin 2 x = 0 .
dx
If y = tan −1 x show that
(1 + x)2 y 2 + 2xy1 = 0 .
10.
If y = (cos −1 x )2 , show that
(1 − x 2 )y 2 − xy1 − 2 = 0 .
274
MATHEMATICS
Differentiation of Trigonometric Functions
MODULE - V
Calculus
ANSWERS
CHECK YOUR PROGRESS 22.1
(1)
2.
(a) − cosec x c o t x
(b) − cosec 2 x
(d) −2cosec2 2x
(e) −2xcosecx 2 c o t x2 (f)
(a) 2sin2x
(b) −2cosec 2 x c o t x
(c) −2sin2x
Notes
cosx
2 sinx
(c) 2tanxsec 2 x
CHECK YOUR PROGRESS 22.2
1.
2.
sec 2 x
2 x
(a) 12cos4x
(b) −5 s i n 5 x
(c)
(e) 2 x c o s x2
(f) 2 2 sec2 2x
(g) − 3π cosec2 3x
(h) 10sec10xtan10x
(i) −2cosec2xcot2x
(a)
2secxtanx
2
(secx + 1)
(b)
−2
(d)
cos x
2 x
(c) xcosx + sinx
(sinx − cosx)2
(d) 2xcosx − (1 + x 2 )sinx
(e) cosecx(1 − xcotx) (f) 2cos2xcos3x − 3 s i n 2 x s i n 3 x
3.
(a) 3sin 2 xcosx
(b) − sin2x
(e) 5sec 5 x t a n x
(f) −3cosec3x cotx
(g)
3cos3x
2 sin3x
(d) −4cot 3 xcosec 2 x
(c) 4tan 3 xsec2 x
(g) sec x tan x
2 x
4.
(h) secx (secx + tanx)
(a) 1
(b) 2 + 2
CHECK YOUR PROGRESS 22.3
1.
(i)
(iv)
−2x
1 − x4
2x
1+ x
(ii)
(v)
4
−1
x 1− x
2
1
x(1 + x 2 )
−
−
− cos− 1 x
x
(iii)
2
tan −1 x
−1
1
2x 2
(vi)
x2
(1 − x)
1
1
2x 2
(1 + x )
CHECK YOUR PROGRESS 22.4
2x
1.
(a)
1− x
MATHEMATICS
4
(b)
−1
4−x
1
2
(c)
x x2 − 1
275
Differentiation of Trigonometric Functions
MODULE - V
2.
Calculus
3.
Notes
(a)
1
2
(a) −
(c)
−
(b)
cos ( cos−1 x)
(b)
1− x 2
−2
1− x
(d)
2
x
1+ x
2
1
2
(c) −1
⋅ sec ( tan −1 x )
−3
1− x
(e)
2
−1
2(1 + x 2 )
1
4.
(1+ tan −1 x)
2
CHECK YOUR PROGRESS 22.5
1.
− cosxcos(cosx) − sin 2 xsin(cosx)
(a)
2x(2 + x 2 )
(b)
2 2
(1 + x )
+ 2tan −1 x
TERMINAL EXERCISE
1.
3.
x
x
x
x 3 tan sec2 + 3x 2 tan2
2
2
2
5(3 − x)
3(1 −
6.
276
5
x)3
1
2y − 1
− 2sin(4x + 2)
2.
0, 0
5.
|sec θ |
1
7.
2 1− x 2
MATHEMATICS