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

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

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)]

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)]

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

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

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)]

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)]

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

Question 13.
y = (2x – 5)⁴ (8x² – 5) ^{– 3}
Answer:

Question 14.
y = (x² + 1) \(\sqrt[3]{x^{2}+2}\)
Answer:

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:

Question 17.
f(x) = \(\frac{x}{\sqrt{7-3 x}}\)
Answer:

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 = \(\frac{\sin ^{2} x}{\cos x}\)

= 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 = \(5^{\frac{-1}{x}}\)

Question 21.
y = \(\sqrt{1+2 \tan x}\)
Answer:

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)

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:

Question 26.
y = \(\sqrt{x+\sqrt{x}}\)
Answer:

[y = f(g(x))
\(\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]

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

Question 28.
y = \(\sqrt{x+\sqrt{x+\sqrt{x}}}\)
Answer:

[y = f(g(x))
\(\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]

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)]

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)]