Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.3

Differentiate the following:

Question 1.
y = (x2 + 4x + 6)5
Answer:
Let = u = x2 + 4x + 6
⇒ \(\frac{d u}{d x}\) = 2x + 4
Now y = u5 ⇒ \(\frac{d y}{d x}\) = 5u4
∴ \(\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}\) = 5u4 (2x + 4)
= 5(x2 + 4x + 6)4 (2x + 4)
= 5 (2x + 4) (x2 + 4x + 6)4

Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3

Question 2.
y = tan 3x
Answer:
y = tan 3x
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = sec2 3x . \(\frac{\mathrm{d}}{\mathrm{d} x}\) (3x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = sec23x × 3
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = 3 sec2 3x

Question 3.
y = cos (tan x)
Answer:
Put u = tan x
\(\frac{d u}{d x}\) = sec2x
Now y = cos u ⇒ \(\frac{d u}{d x}\) = -sin u
Now \(\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}\)
= (-sin u) (sec2x)
= -sec2 (sin (tan x))

Question 4.
y = \(\sqrt[3]{1+x^{3}}\)
Answer:
y = \(\sqrt[3]{1+x^{3}}\)
y = (1 + x3)1/3
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 1

Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3

Question 5.
y = \(\mathrm{e}^{\sqrt{x}}\)
Answer:
y = \(\mathrm{e}^{\sqrt{x}}\)
y = \(e^{x^{\frac{1}{2}}}\)
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 2

Question 6.
y = sin (ex)
Answer:
y = sin (ex)
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]
y = cos (ex) . \(\frac{\mathrm{d}}{\mathrm{d} x}\) (ex)
y = cos ((ex)) . ex
y = ex cos (ex)

Question 7.
F(x) = (x3 + 4x)7
Answer:
F(x) = (x3 + 4x)7
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]
F’ (x) = 7 (x3 + 4x)7 – 1 \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x3 + 4x)
= 7 (x3 + 4x)6 (3x2 + 4)
= 7 (3x2 + 4) (x3 + 4x)6

Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3

Question 8.
h (t) = \(\left(t-\frac{1}{t}\right)^{\frac{3}{2}}\)
Answer:
h (t) = \(\left(t-\frac{1}{t}\right)^{\frac{3}{2}}\)
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 3

Question 9.
f(t) = \(\sqrt[3]{1+\tan t}\)
Answer:
f(t) = \(\sqrt[3]{1+\tan t}\)
f(t) = (1 + tan t)1/3
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 4

Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3

Question 10.
y = cos (a3 + x3)
Answer:
y = cos (a3 + x3)
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = – sin (a3 + x3) \(\frac{\mathrm{d}}{\mathrm{d} x}\) (a3 + x3)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = – sin (a3 + x3) (0 + 3x2)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = – 3x2 sin (a3 + x3)

Question 11.
y = e-mx
Answer:
y = e-mx
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = e-mx × \(\frac{\mathrm{d}}{\mathrm{d} x}\) (- mx)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = e-mx × – m
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = – m e-mx = – my

Question 12.
y = 4 sec 5x
Answer:
y = 4 sec 5x
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = 4 × sec 5x × tan 5x \(\frac{\mathrm{d}}{\mathrm{d} x}\) (5x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = 4 sec 5x tan 5x × 5 × 1
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = 20 sec 5x tan 5x

Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3

Question 13.
y = (2x – 5)4 (8x2 – 5) – 3
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 5
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 6

Question 14.
y = (x2 + 1) \(\sqrt[3]{x^{2}+2}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 7

Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3

Question 15.
y = x e-x2
Answer:
y = xe-x2
y = uv where u = x and v = e-x2
Now u’ = 1 and v’ = e-x2 (-2x)
v’ = – 2xe-x2
Now y = uv ⇒ y’ = uv’ + vu’
(i.e.) \(\frac{d y}{d x}\) = x[-2xe-x2] + e-x2 (1)
= e-x2 (1 – 2x2)

Question 16.
s(t) = \(\sqrt[4]{\frac{t^{3}+1}{t^{3}-1}}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 8

Question 17.
f(x) = \(\frac{x}{\sqrt{7-3 x}}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 9

Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3

Question 18.
y = tan (cos x)
Answer:
y = tan (cos x)
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = sec2 (cos x) × \(\frac{\mathrm{d}}{\mathrm{d} x}\) (cos x)
= sec2 (cos x) × – sin x
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = -sin x . sec2 (cos x)

Question 19.
y = \(\frac{\sin ^{2} x}{\cos x}\)
Answer:
y = \(\frac{\sin ^{2} x}{\cos x}\)
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 10
= sin x (2 + tan2x)
= sin x (1 + 1 + tan2x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = sin x (1 + sec2 x)

Question 20.
y = \(5^{\frac{-1}{x}}\)
Answer:
y = \(5^{\frac{-1}{x}}\)
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 11

Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3

Question 21.
y = \(\sqrt{1+2 \tan x}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 12

Question 22.
y = sin3x + cos3x
Answer:
y = sin3x + cos3x
Here u = sin3 x = (sin x)3
⇒ \(\frac{d u}{d x}\) = 3 (sin x)2 (cos x)
= 3sin2x cos x
v = cos3x = (cos x)3
⇒ \(\frac{d v}{d x}\) = 3 (cos x)2 (-sin x) = -3 sin x cos2x
Now y = u + v ⇒ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
= 3 sin2x cos x – 3sin x cos2x
= 3 sin x cos x (sin x – cos x)

Question 23.
y = sin2 (cos kx)
Answer:
y = sin2 (cos kx)
y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = 2 sin (cos kx) × cos (cos kx) × -sin kx × k × 1
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = sin (2 cos kx) × -k sin kx
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = -k sin kx . sin (2 cos kx)

Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3

Question 24.
y = (1 + cos2)6
Answer:
y = (1 + cos2)6
y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = 6(1 + cos2x)6-1 (0 + 2 cos x × -sin x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = 6(1 + cos2x)5 × – 2 sin x cos x
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = -6 sin 2x (1 + cos2)5

Question 25.
y = \(\frac{e^{3 x}}{1+e^{x}}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 13
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 14

Question 26.
y = \(\sqrt{x+\sqrt{x}}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 15
[y = f(g(x))
\(\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 16

Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3

Question 27.
y = ex cos x
Answer:
y = ex cos x
\(\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}\) = ex cos x (x – sinx + cos x . 1)
\(\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}\) = ex cos x (cos x – x sin x)

Question 28.
y = \(\sqrt{x+\sqrt{x+\sqrt{x}}}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 17
[y = f(g(x))
\(\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 18Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 19

Question 29.
y = sin (tan (\(\sqrt{\sin x}\)))
Answer:
y = sin (tan (\(\sqrt{\sin x}\)))
y = f(g(x))
\(\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 20

Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3

Question 30.
y = sin-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Answer:
y = sin-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
[y = f(g(x))
\(\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]
Samacheer Kalvi 11th Maths Guide Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 21

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