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

## 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

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:

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 = $$\frac{u}{v} \Rightarrow y^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^{2}}$$

Question 10.
y = $$\frac{x}{\sin x+\cos x}$$
Solution:

Question 11.
y = $$\frac{\tan x-1}{\sec x}$$
Solution:

Question 12.
y = $$\frac{\sin x}{x^{2}}$$
Solution:

Question 13.
y = tan θ (sin θ + cos θ)
Solution:

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:

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:

Question 18.
y = sin x0
Solution:

Question 19.
y = log10x
Solution:

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 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:

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

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

Question 9.
$$\frac{\tan x+1}{\tan x-1}$$
Solution:

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’ = $$\frac{v u^{\prime}-u v^{\prime}}{v^{2}}$$