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

## 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
Solution:
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
Solution:
y = tan 3x
put u = 3x
$$\frac{d u}{d x}$$ = 3
Now y = tan u
⇒ $$\frac{d u}{d x}$$ = sec2 u
So $$\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}$$ = (sec2 u) (3)
= 3 sec2 3x

Question 3.
y = cos (tan x)
Solution:
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 = $$\sqrt[3]{1+x^{3}}$$
Solution:

Question 5.
y = $$e^{\sqrt{x}}$$
Solution:

Question 6.
y = sin (ex)
Solution:
y = sin (ex)
Let u = ex

Question 7.
F(x) = (x3 + 4x)7
Solution:
F(x) = (x3 + 4x)7
Put u = x3 + 4x

Question 8.
h(t) = $$\left(t-\frac{1}{t}\right)^{\frac{3}{2}}$$
Solution:

Question 9.
f(t) = $$\sqrt[3]{1+\tan t}$$
Solution:

Question 10.
y = cos (a3 + x3)
Solution:

Question 11.
y = e-mx
Solution:

Question 12.
y = 4 sec 5x
Solution:

Question 13.
y = (2x – 5)4 (8x2 – 5)-3
Solution:

= $$\frac{8(2 x-5)^{3}}{\left(8 x^{2}-5\right)^{4}}$$ (-4x2 + 30x – 5)

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

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

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

Question 18.
y = tan (cos x)
Solution:
y = tan (cos x)

Question 19.
y = $$\frac{\sin ^{2} x}{\cos x}$$
Solution:

Question 20.
y = $$5^{-\frac{1}{x}}$$
Solution:

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

Question 22.
y = sin3x + cos3x
Solution:
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)
Solution:

Question 24.
y = (1 + cos2x)6
Solution:

Question 25.
y =$$\frac{e^{3 x}}{1+e^{x}}$$
Solution:

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

Question 27.
y = ex cos x
Solution:

Question 28.
y = $$\sqrt{x+\sqrt{x+\sqrt{x}}}$$
Solution:

Question 29.
y = $$\sin (\tan (\sqrt{\sin x}))$$
Solution:

Question 30.
y = sin-1$$\left(\frac{1-x^{2}}{1+x^{2}}\right)$$
Solution: