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
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
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)
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 = $$\sqrt[3]{1+x^{3}}$$
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 = $$\mathrm{e}^{\sqrt{x}}$$
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 (ex)
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
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) = $$\left(t-\frac{1}{t}\right)^{\frac{3}{2}}$$
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) = $$\sqrt[3]{1+\tan t}$$
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)
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
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
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

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

Question 15.
y = x e-x2
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) = $$\sqrt[4]{\frac{t^{3}+1}{t^{3}-1}}$$

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

Question 18.
y = tan (cos x)
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 = $$\frac{\sin ^{2} x}{\cos x}$$
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 = $$5^{\frac{-1}{x}}$$
y = $$5^{\frac{-1}{x}}$$

Question 21.
y = $$\sqrt{1+2 \tan x}$$

Question 22.
y = sin3x + cos3x
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)
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
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 = $$\frac{e^{3 x}}{1+e^{x}}$$

Question 26.
y = $$\sqrt{x+\sqrt{x}}$$

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

Question 27.
y = ex cos x
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 = $$\sqrt{x+\sqrt{x+\sqrt{x}}}$$

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

Question 29.
y = sin (tan ($$\sqrt{\sin x}$$))
y = sin (tan ($$\sqrt{\sin x}$$))
$$\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}$$ = f'(g(x)) . g'(x)]
y = sin-1$$\left(\frac{1-x^{2}}{1+x^{2}}\right)$$
y = sin-1$$\left(\frac{1-x^{2}}{1+x^{2}}\right)$$
$$\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}$$ = f'(g(x)) . g'(x)]