TS Inter 1st Year Maths 1B Solutions Chapter 9 Differentiation Ex 9(d)

Students must practice these TS Inter 1st Year Maths 1B Study Material Chapter 9 Differentiation Ex 9(d) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Differentiation 9(d)

I.
Question 1.
If y = \(\frac{2 x+3}{4 x+5}\) then find y”. (V.S.A.Q.)
Answer:
TS Inter 1st Year Maths 1B Solutions Chapter 9 Differentiation Ex 9(d) 1

TS Inter 1st Year Maths 1B Solutions Chapter 9 Differentiation Ex 9(d)

Question 2.
If y = aenx + be-nx then prove that y” = n2y. (May 2014) (V.S.A.Q.)
Answer:
y’ = anenx – bne-nx
y” = an2enx + bn2e-nx
= n2(aenx + be– nx) = n2y
∴ y” = n2y

II.
Question 1.
Find the second order derivatives of the following functions f(x). (S.A.Q.)
(i) cos3x
Answer:
Let y = cos3x = \(\frac{1}{4}\)[cos 3x + 3 cos x]
y’ = \(\frac{1}{4}\) [- 3 sin x – 3 sin x]
y” = \(\frac{1}{4}\) [- 9 cos 3x – 3 cos x]
= – \(\frac{3}{4}\) (3 cos 3x + cos x)

(ii) sin4x
Answer:
Let y = sin4x sin3x sin x
TS Inter 1st Year Maths 1B Solutions Chapter 9 Differentiation Ex 9(d) 2
y” = 2 cos 2x – 2 cos 4x
= 2 (cos 2x – cos 4x)

(iii) log (4x2 – 9)
Answer:
Let y = log (4x2 – 9)
= log [(2x – 3) (2x + 3)]
= log (2x + 3) + log (2x + 3)
TS Inter 1st Year Maths 1B Solutions Chapter 9 Differentiation Ex 9(d) 3

TS Inter 1st Year Maths 1B Solutions Chapter 9 Differentiation Ex 9(d)

(iv) e-2x sin3x
Answer:
Let y = e-2x sin3x
y ‘ = e-2x (3 sin2x cos x) + sin3x (-2) e-2x
= e-2x [ 3 sin2x cos x – 2 sin3x]
y” = e-2x [3 sin2x (- sin x) + cos x (6 sin x cos x) – 6 sin2x cos x] – 2e-2x [ 3 sin2x cos x – 2 sin3x]
= – e-2x [6 sin x cos2x – 6 sin2x cos x – 3 sin3x – 6 sin2x cos x + 4 sin3x]
= e-2x [sin3x – 12 sin2x cos x + 6 sin x cos2x]

(v) ex sin x cos 2x
Answer:
Let y = ex sin x cos 2x
= ex \(\frac{1}{2}\) [sin 3x – sin x]
y’ = \(\frac{1}{2}\)[ex (3 cos 3x – cos x) + ex (sin 3x – sin x)]
y” = \(\frac{1}{2}\)[ex (- 9 sin 3x + sin x) + ex (3 cos 3x – cos x) + ex (3 cos 3x – cos x) + (sin 3x – sin x)ex]
= \(\frac{1}{2}\) ex [6 cos 3x – 8 sin 3x – 2 cos x]
= ex [3 cos 3x – 4 sin 3x – cos x]

(vi) tan-1 \(\left(\frac{1+x}{1-x}\right)\)
Answer:
TS Inter 1st Year Maths 1B Solutions Chapter 9 Differentiation Ex 9(d) 4

(vii) tan-1\(\left(\frac{3 x-x^3}{1-3 x^2}\right)\)
Answer:
TS Inter 1st Year Maths 1B Solutions Chapter 9 Differentiation Ex 9(d) 5

Question 2.
Prove the following. (S.A.Q.)
(i) If y = axn + 1 + bx-n then x2y” = n(n + 1) y.
Answer:
y = axn + 1 + bx-n
y’ = (n + 1) ax-n – nbx– n – 1
y” = an (n + 1) xn + 1 + bn (n + 1) x-n-2
x2y” = an(n + 1) xn + 1 + b(n + 1)n x– n
= n(n + 1) [axn + 1 + bx– n]
= n(n + 1) y

TS Inter 1st Year Maths 1B Solutions Chapter 9 Differentiation Ex 9(d)

(ii) If y = a cos x + (b + 2x) sin x then y” + y = 4 cos x.
Answer:
y = a cos x + (b + 2x) sin x
y’ = – a sin x + (b + 2x) cos x + 2 sin x
y” = – a cos x + (b + 2x) (- sin x) + 2 cos x + 2 cos x
= – a cos x + 4 cos x – (b + 2x) sin x
= -[a cos x + (b + 2x) sin x] + 4 cos x
= – y + 4 cos x
∴ y” + y = 4 cos x

(iii) If y = 6 (x + 1) + (a + bx) e3x then y” – 6y’ + 9y = 54x + 18
Answer:
y = 6 (x + 1) + (a + bx) e3x
∴ y’ = 6 + (a + bx) 3e3x + be3x
y” = (a + bx) 9e3x + 3e3x(b) + 3be3x
∴ L.H.S. = y” – 6y’ + 9y
= (a + bx) 9e3x + 6be3x – 6 [6 + (a + bx) 3e3x + be3x] + 9 [ 6 (x + 1) + (a + bx) e3x]
= (a + bx) 9e3x + ebe3x – 36 – 18e3x (a + bx) – 6be3x+ 54 (x + 1) + 9 (a + bx) e3x
= 54x+ 18

(iv) If ay4 = (x + b)5 then 5yy” = (y’)2
Answer:
TS Inter 1st Year Maths 1B Solutions Chapter 9 Differentiation Ex 9(d) 6

(v) If y = a cos (sin x) + b sin (sin x) then y” + (tan x)y’ + y cos2x = 0
Answer:
Given y = a cos (sin x) + b sin (sin x) then
y’ = -a sin (sin x) cos x + b cos (sin x) cos x
= cos x [-a sin (sin x) + b cos (sin x)]
y” = cos x [- a cos (sin x) cos x – b sin (sin x) (cos x)] – sin x [- a sin (sin x) + b cos (sin x)]
= – sin x \(\frac{y^{\prime}}{\cos x}\) – cos2 x . y
∴ y” + (tan x) y’ + y cos2x = 0

III.
Question 1.
(i) If y = 128 sin3 x cos4x then find y”. (E.Q.)
Answer:
Given y = 128 sin3x cos4x
∴ y’ = 128 [sin3x 4 cos3x (- sin x) + cos4x (3 sin2x) cos x]
= 128 [3 sin2x cos5x – 4 sin4x cos3x]
y” = 128[3 sin2x (5 cos4x) (- sin x) + cos5x (6 sin x cos x) – 4 [sin4x (3 cos2x) (- sin x) + c0s3x 4 sin3x cos x]]
= 128 [- 15 sin3 x cos4x + 6 cos6x sin x + 12 sin5x cos2x – 16 cos4x sin3x]
= 128 [6 sin x cos6x + 12 sin5x cos2x – 31 cos4x sin3x]
= 128 sin x cos2x (12 sin4x – 31 sin2x cos2x + 6 cos4x)

TS Inter 1st Year Maths 1B Solutions Chapter 9 Differentiation Ex 9(d)

(ii) If y = sin 2x sin 3x sin 4x then find y”. (E.Q.)
Answer:
y = sin 2x \(\frac{1}{2}\) [2sin 3x sin 4x]
= sin 2x \(\frac{1}{2}\) [cos x – cos 7x]
= \(\frac{1}{2}\) sin 2x cos x – \(\frac{1}{2}\) sin 2x cos 7x
= \(\frac{1}{4}\) [sin 3x + sin x – sin 9x + sin 5x]
= \(\frac{1}{4}\) [- sin 9x + sin 5x + sin 3x + sin x]
y’ = \(\frac{1}{4}\) [- 9 cos 9x + 5 cos 5x + 3 cos 3x + cos x] 1
y” = \(\frac{1}{4}\) [81 sin 9x – 25 sin 5x – 9 sin 3x – sin x]

(iii) If ax2 + 2hxy + by2 = 1 then prove that \(\frac{d^2 y}{d x^2}=\frac{h^2-a b}{(h x+b y)^3}\) (E.Q.)
Answer:
Differentiating ax2 + 2hxy + by2 = 1 with respect to x on both sides we get
TS Inter 1st Year Maths 1B Solutions Chapter 9 Differentiation Ex 9(d) 7

(iv) If y = ae-bx cos (cx + d) then prove that y” + 2by’ + (b2 + c2)y = 0 (E.Q.)
Answer:
Given y = ae-bx cos (cx + d)
Differentiating w.r.t ‘x’ then
y’ = a[e-bx(-c sin (cx + d)) – cos(cx + d) be-bx]
= ae-bx [- c sin(cx + d) – bcos(cx + d)]
= – ace-bx sin (cx + d) – by
∴ y’ + by = -ace-bx sin(cx + d)
Differentiating again w.r.t ‘x’
∴ y” + by’ = -ac [e-bx c cos (cx + d) – be-bx sin (cx + d)]
= – ac e-bx [c cos (cx + d) – b sin(cx + d)]
= abce-bx sin (cx + d) – ac2e-bxcos (cx + d)
= – b (y’ + by) – c2y
∴ y” + 2by’ + (b2 + c2)y = 0

TS Inter 1st Year Maths 1B Solutions Chapter 9 Differentiation Ex 9(d)

(v) If y = e–\(\frac{\mathbf{k}}{2}\)x (a cos nx + b sin nx) then prove that y” + ky’ + (n2 + \(\frac{k^2}{4}\)) y = 0 (E.Q.)
Answer:
TS Inter 1st Year Maths 1B Solutions Chapter 9 Differentiation Ex 9(d) 8

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