CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Ex 7(c)

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 7 Continuity and Differentiability Ex 7(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Exercise 7(c)

Find derivatives of the following functions.
Question 1.
(x2 +5)8
Solution:
y = (x2 +5)8
$$\frac{d y}{d x}$$ = $$\frac{d}{d x}$$(x2 +5)8
= 8(x2 +5)7 × $$\frac{d}{d x}$$(x2 +5) by chain rule
= 8(x2 +5)7 . 2x
= 16x (x2 +5)7

Question 2.
$$\frac{1}{\left(x^3+\sin x\right)^2}$$
Solution:

Question 3.
In (√x+1)
Solution:

Question 4.
sin 5x + cos 7x
Solution:
sin 5x + cos 7x
$$\frac{d y}{d x}$$ = $$\frac{d}{d x}$$(sin 5x) + $$\frac{d}{d x}$$(cos 7x)
= cos 5x . $$\frac{d}{d x}$$(5x) – sin 7x . $$\frac{d}{d x}$$(7x)
= 5 cos 5x – 7 sin 7x

Question 5.
esin t
Solution:
y = esin t
$$\frac{d y}{d x}$$ = $$\frac{d}{d t}$$(esin t) = esin t . $$\frac{d}{d t}$$(sin t)
= esin t . cos t

Question 6.
$$\sqrt{a x^2+b x+c}$$
Solution:
y = $$\sqrt{a x^2+b x+c}$$

Question 7.
$$\left(\frac{x+1}{x^2+3}\right)^{-3}$$
Solution:

Question 8.
sec (tan θ)
Solution:
y = sec (tan θ)
$$\frac{d y}{d θ}$$ = $$\frac{d}{d θ}$$ {sec (tan θ)}
= sec (tan θ) . tan (tan θ) . $$\frac{d}{d θ}$$(tan θ)
= sec (tan θ) . tan (tan θ) . sec2 θ

Question 9.
sin $$\left(\frac{1-x^2}{1+x^2}\right)$$
Solution:

Question 10.
$$\sqrt{\tan (3 z)}$$
Solution:

Question 11.
tan3 x
Solution:
y = tan3x = (tan x)3
$$\frac{d y}{d x}$$ = 3(tan x)2. $$\frac{d}{d x}$$(tan x)
= 3tan2 x . sec2 x

Question 12.
sin4 x
Solution:
y = sin4x
$$\frac{d y}{d x}$$ = $$\frac{d}{d x}$$(sin4 x)
= 4 (sin x)3 . $$\frac{d}{d x}$$(sin x)
= 4 sin3 x . cos x

Question 13.
sin2 x cos2 x
Solution:
y = sin2 x cos2 x = $$\frac{1}{4}$$sin2 2x
$$\frac{d y}{d x}$$ = $$\frac{1}{4}$$ $$\frac{d}{d x}$$(sin 2x)2
= $$\frac{1}{4}$$. 2sin 2x . $$\frac{d}{d x}$$(sin 2x)
= $$\frac{1}{2}$$ sin 2x . cos 2x . $$\frac{d}{d x}$$(2x)
= $$\frac{1}{2}$$ sin 2x cos 2x . 2 = sin 2x . cos 2x

Question 14.
sin 5x cos 7x
Solution:
y = sin 5x cos 7x
$$\frac{d y}{d x}$$ = $$\frac{d}{d x}$$(sin 5x) . cos 7x + $$\frac{d}{d x}$$sin 5x . (cos 7x)
= cos 5x . $$\frac{d}{d x}$$(5x) . cos 7x + sin 5x . (-sin 7x) . $$\frac{d}{d x}$$(7x)
= 5 cos 5x . cos 7x – 7 sin 5x . sin 7x

Question 15.
tan x cot 2x
Solution:
y = tan x. cot 2x
$$\frac{d y}{d x}$$ = $$\frac{d}{d x}$$(tan x) . cot 2x + tan x . $$\frac{d}{d x}$$(cot 2x)
= sec2 x / cot 2x + tan x . (-cosec2 x) $$\frac{d}{d x}$$(2x)
= sec2 x . cot 2x – 2 tan x . cosec2 x

Question 16.
$$\sqrt{\sin \sqrt{x}}$$
Solution:

Question 17.
$$\sqrt{\sec (2 x+1)}$$
Solution:

Question 18.
cosec (ax + b)2
Solution:
y = cosec (ax + b)2
$$\frac{d y}{d x}$$ = – cosec (ax + b)2 cot (ax + b)2 . $$\frac{d}{d x}$$(ax + b)2
[ ∵ $$\frac{d}{d x}$$(cosec u) = -cosec u . cot u . $$\frac{d u}{d x}$$
= – cosec (ax + b)2 . cot (ax + b)2 . 2(ax + b) . $$\frac{d}{d x}$$(ax + b)
[ ∵ $$\frac{d}{d x}$$(u2) = 2u $$\frac{d u}{d x}$$
= – cosec (ax + b)2 . cot (ax + b)2 . 2(ax + b) . a
= – 2a (ax + b) cosec (ax + b)2 cot (ax + b)2

Question 19.
aIn x
Solution:
y = aIn x . In a . $$\frac{d}{d x}$$(In x)
[ ∵ $$\frac{d}{d x}$$(au) = au . In a . $$\frac{d u}{d x}$$
= aIn x . In a . $$\frac{1}{x}$$ = $$\frac{a^{\ln x} \ln a}{x}$$

Question 20.
$$a^{x^2} b^{x^3}$$
Solution:

Question 21.
In tan x
Solution:

Question 22.
$$5^{\sin x^2}$$
Solution:

Question 23.
In tan$$\left(\frac{\pi}{4}+\frac{x}{2}\right)$$
Solution:

Question 24.
$$\sqrt{\left(a^{\sqrt{x}}\right)}$$
Solution:

Question 25.
In (enx + e-nx)
Solution:
y = In (enx + e-nx)
$$\frac{d y}{d x}$$ = $$\frac{1}{e^{n x}+e^{-n x}}$$ . $$\frac{d}{d x}$$ (enx + e-nx)
= $$\frac{n\left(e^{n x}-e^{-n x}\right)}{e^{n x}+e^{-n x}}$$

Question 26.
$$e^{\sqrt{a x}}$$
Solution:

Question 27.
$$\sqrt{\log x}$$
Solution:
y = $$\sqrt{\log x}$$
$$\frac{d y}{d x}$$ = $$\frac{1}{2 \sqrt{\log x}}$$ . $$\frac{d}{d x}$$(log x)
= $$\frac{1}{2 \sqrt{\log x}}$$ . $$\frac{1}{x}$$

Question 28.
esin x – acos x
Solution:
y = esin x – acos x
$$\frac{d y}{d x}$$ = $$\frac{d}{d x}$$(esin x) – $$\frac{d}{d x}$$(acos x)
= esin x . $$\frac{d}{d x}$$(sin x) – acos x . In a . $$\frac{d}{d x}$$(cos x)
= esin x . cos x + acos x . In a . sin x

Question 29.
$$\frac{e^{3 x^2}}{\ln \sin x}$$
Solution:

Question 30.
Prove that
$$\frac{d}{d x}\left[\frac{1-\tan x}{1+\tan x}\right]^{\frac{1}{2}}$$ = 1 / $$\sqrt{\cos 2 x}$$ (cos x + sin x)
Solution: