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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2

Find the derivatives of the following functions with respect to corresponding independent variables.
Question 1.
f(x) = x – 3 sinx
Solution:
f(x) = x – 3 sinx
= f'(x) = 1 – 3 (cos x)
= 1 – 3 cos x

Question 2.
y = sin x + cos x
Solution:
\(\frac{d y}{d x}\) = cosx + (-sinx) = cos x – sin x

Question 3.
f(x) = x sin x
Solution:
f(x) = uv
⇒ f'(x) = uv’ + vu’ = u\(\frac{d u}{d x}\) + v\(\frac{d v}{d x}\)
Now u = x ⇒ u’ = 1
v = sin x ⇒ v’ cos x
f'(x) = x (cos x) + sin x(1)
= x cos x + sin x

Question 4.
y = cos x – 2 tan x
Solution:
\(\frac{d y}{d x}\) = -sin x = 2 (sec2x)
= – sin x – 2 sec2x

Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2

Question 5.
g(t) = t3 cos t
Solution:
g(t) = t3 cost (i.e.) u = t3 and v = cos t
let u’ = \(\frac{d u}{d x}\) and v’= \(\frac{d v}{d x}\) = (-sint)
g'(t) = uv’ + vu’
g'(t) = t3 (-sin t) + cos t (3t2)
= -t3 sin t + 3t2 cos t

Question 6.
g(t) = 4 sec t + tan t
Solution:
g{t) = 4 sect + tan t
g'(t) = 4(sec t tan t) + sec2t
= 4sec t tan t + sec2t

Question 7.
y = ex sin x
Solution:
y = ex sin x
⇒ y = uv’ + vu’
Now u = ex ⇒ u’ = \(\frac{d u}{d x}\) ex
v = sin x ⇒ v’ = \(\frac{d v}{d x}\) cos x
i.e. y’ = ex (cos x) + sin x (ex)
= ex [sin x + cos x]

Question 8.
y = \(\frac{\tan x}{x}\)
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 1

Question 9.
y = \(\frac{\sin x}{1+\cos x}\)
Solution:
y = \(\frac{\sin x}{1+\cos x}=\frac{u}{v}\) (say)
u = sin x v = 1 + cosx
u’ = cos x v’ = -sin x
y = \(\frac{u}{v} \Rightarrow y^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^{2}}\)
Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 2

Question 10.
y = \(\frac{x}{\sin x+\cos x}\)
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 3

Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2

Question 11.
y = \(\frac{\tan x-1}{\sec x}\)
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 4

Question 12.
y = \(\frac{\sin x}{x^{2}}\)
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 5

Question 13.
y = tan θ (sin θ + cos θ)
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 6

Question 14.
y = cosex x. cot x
Solution:
y = u v ⇒ y’ = uv’ + vu’
u = cosec x ⇒ u’ = -cosec x cot x
v = cot x ⇒ v’ = – cosec2 x
(cosec x)(-cosec2x) + cot x(-coseç x cot x)
= cosec3x – cosec x cot2x
= – cosec x (cosec2x + cot2x)
= \(-\frac{1}{\sin x}\left(\frac{1+\cos ^{2} x}{\sin ^{2} x}\right)=-\frac{\left(1+\cos ^{2} x\right)}{\sin ^{3} x}\)

Question 15.
y = x sin x cos x
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 7

Question 16.
y = e-x. log x
Solution:
y = e-x logx = uv (say)
Here u = e-x and v = log x
⇒ u’ = -e-x and v’ = \(\frac{1}{x}\)
Now y = uv ⇒ y’ = uv’ + vu’
(i.e.) \(\frac{d y}{d x}\) = e-x \(\left(\frac{1}{x}\right)\) + log x(-e-x)
= e-x(\(\frac{1}{x}\) – log x)

Question 17.
y = (x2 + 5) log (1 + x)e-3x
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 8
Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 9

Question 18.
y = sin x0
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 10

Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2

Question 19.
y = log10x
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 11

Question 20.
Draw the function f'(x) if f(x) = 2x2 – 5x + 3
Solution:
f(x) = 2x2 – 5x + 3
f'(x) = 4x – 5 which is a linear function
(i.e.) y = 4x – 5
Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 12

Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 Additional Questions

Question 1.
Find the derivation of following functions
Question 1.
3 sin x + 4 cos x – ex
Solution:
y = 3 sin x + 4 cos x – ex
\(\frac{d y}{d x}\) = 3 (cos x) + 4 (- sin x) – (ex)
= 3 cos x – 4 sin x – ex

Question 2.
sin 5 + log10x + 2 secx
Solution:
y = sin 5 + log10x + 2 secx
\(\frac{d y}{d x}\) = 0 + \(\left(\frac{1}{x}\right)\) log10 e + 2[sec x + tan x] = \(\frac{\log _{10} e}{x}\) + 2 sec x tan x

Question 3.
6 sin x log10x + e
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 13

Question 4.
(x4 – 6x3 + 7x2 + 4x + 2) (x3 – 1)
Solution:
Let u = x4 – 6x3 + 7x2 + 4x + 2 and v = x3 – 1
u’ = 4x3 – 6 (3x2) + 7 (2x) + 4 (1) + 0
= 4x3 – 18x2 + 14x + 4
v’= 3x3
y = uv’ + vu’
i.e. \(\frac{d y}{d x}\) = (x4 – 6x3 + 7x2 + 4x + 2) (3x2) + (x3 – 1) (4x3 – 18x2 + 14x + 4)
= 3x6 – 18x5 + 21x4 + 12x3 + 6x2 + 4x6 – 18x5 + 14x4 + 4x3 – 4x3 + 18x2 – 14x – 4
= 7x6 – 36x5 + 35x4 + 12x3 + 24x2 – 14x – 4

Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2

Question 5.
(3x2 + 1)2
Solution:
y = (3x2 + 1)2 = (3x2 + 1) (3x2 + 1)
Let u = 3x2 + 1 and v = 3x2 + 1
∴ u’ = 3(2x) = 6x and v’ = 6x
y’ = uv’ + vu’
(i.e.,) \(\frac{d y}{d x}\) = (3x2 + 1) (6x) + (3x2 + 1) 6x = 12x (3x2 + 1)

Question 6.
(3 sec x – 4 cosec x) (2 sin x + 5 cos x)
Solution:
y = (3 sec x – 4 cosec x) (2 sin x + 5cos x)
Let u = 3 secx-4 cosecx and v = 2 sinx + 5 cosx
u’ = 3 (sec x tan x) – 4 (-cosec x cot x) ; v’ = 2 (cos x) + 5 (- sin x)
u’ = 3 sec x tan x + 4 cosec x cot x); v’ = 2 cos x – 5 sin x .
∴ y’ = uv’ + vu’
So \(\frac{d y}{d x}\) = (3 sec x – 4 cosec x) (2 cos x – 5 sinx) + (2 sin x + 5 cos x) (3 sec x tan x + 4 cosec x cot x) = 6 sec x cos x – 15 sec x sin x – 8 cosec x cos x + 20 cosec x sin x + 6 sinx secx tanx + 8 sinx cosecx cotx+ 15 cosx secx tanx + 20 cos x cosec x cot x
= 6 \(\frac{1}{\cos x}\) cosx – 15 \(\frac{1}{\cos x}\)sin x – 8 \(\frac{1}{\sin x}\) cos x + 20 \(\frac{1}{\sin x}\) sin x + 6 sin x \(\frac{1}{\cos x}\) tan x + 8 sin x \(\frac{1}{\sin x}\) cot x + 15 cos x \(\frac{1}{\cos x}\) tan x + 20 cos x \(\frac{1}{\sin x}\) cot x
= 6 – 15tan x – 8cot x + 20 + 6 tan2x + 8 cot x + 15 tan x + 20cot2x
= 26 + 6 tan2x + 20 cot2x

Question 7.
x2 ex sinx
Solution:
y = x2 ex sin x
Let u = x2, v = ex and w = sinx
u’ = 2x, v’ = ex and w’ = cos x
y’ = uvw’ + vwu’ + uwv’
= (x2 ex) cos x + (ex sin x)(2x) + (x2 sin x)ex
= x2 ex cos x + 2xex sin x + x2 ex sin x
= xex {x cos x + 2 sin x + x sin x}

Question 8.
\(\frac{\cos x+\log x}{x^{2}+e^{x}}\)
Solution:
y = \(\frac{\cos x+\log x}{x^{2}+e^{x}}\)
Let u = cos x + log x and v = x2 + ex
∴ u’ = – sin x + \(\frac{1}{x}\), v’ = 2x + ex
Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 14

Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2

Question 9.
\(\frac{\tan x+1}{\tan x-1}\)
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 15

Question 10.
\(\frac{\sin x+x \cos x}{x \sin x-\cos x}\)
Solution:
y = \(\frac{\sin x+x \cos x}{x \sin x-\cos x}\)
Let u = sinx + x cosx and v = x sin x – cos x
u’ = cos x + x( – sin x) + cos x (1)
= cos x – x sin x + cos x = 2 cos x – x sin x
v’=x (cos x) + sin x(1) – (- sin x)
= x cos x + sin x + sin x = 2 sin x + x cos x
y = \(\frac{u}{v}\) ∴ y’ = \(\frac{v u^{\prime}-u v^{\prime}}{v^{2}}\)
Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 16

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