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


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


(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


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


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


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


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


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,


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


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


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


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


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


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


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


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


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


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


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


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


(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


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


− 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