NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5.

• Continuity and Differentiability Class 12 Ex 5.1
• Continuity and Differentiability Class 12 Ex 5.2
• Continuity and Differentiability Class 12 Ex 5.3
• Continuity and Differentiability Class 12 Ex 5.4
• Continuity and Differentiability Class 12 Ex 5.6
• Continuity and Differentiability Class 12 Ex 5.7
• Continuity and Differentiability Class 12 Ex 5.8

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Maths
Chapter	Chapter 5
Chapter Name	Continuity and Differentiability
Exercise	Ex 5.5
Number of Questions Solved	18
Category	NCERT Solutions

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exc  5.5

Differentiate the functions given in Questions 1 to 11 w.r.to x

Question 1.
cos x. cos 2x. cos 3x
Solution:
Let y = cos x. cos 2x . cos 3x,
Taking log on both sides,
log y = log (cos x. cos 2x. cos 3x)
log y = log cos x + log cos 2x + log cos 3x,
Differentiating w.r.t. x, we get

Question 2.

Solution:

taking log on both sides
log y = log

Question 3.
(log x)^cosx
Solution:
let y = (log x)^cosx
Taking log on both sides,
log y = log (log x)^cosx
log y = cos x log (log x),
Differentiating w.r.t. x,

Question 4.
x – 2^sinx
Solution:
let y = x – 2^sinx,
y = u – v

Question 5.
(x+3)².(x + 4)³.(x + 5)⁴
Solution:
let y = (x + 3)².(x + 4)³.(x + 5)⁴
Taking log on both side,
logy = log [(x + 3)² • (x + 4)³ • (x + 5)⁴]
= log (x + 3)² + log (x + 4)³ + log (x + 5)⁴
log y = 2 log (x + 3) + 3 log (x + 4) + 4 log (x + 5)
Differentiating w.r.t. x, we get

Question 6.

Solution:
let
let

Question 7.
(log x)^x + x^logx
Solution:
let y = (log x)^x + x^logx = u+v
where u = (log x)^x
∴ log u = x log(log x)

Question 8.
(sin x)^x+sin^-1 √x
Solution:
Let y = (sin x)^x + sin^-1 √x
let u = (sin x)x, v = sin^-1 √x

Question 9.
x^sinx + (sin x)^cosx
Solution:
let y = x^sinx + (sin x)^cosx = u+v
where u = x^sinx
log u = sin x log x

Question 10.

Solution:

y = u + v

Question 11.

Solution:

Let u = (x cosx)^x
logu = x log(x cosx)

Find of the functions given in Questions 12 to 15.

Question 12.
x^y + y^x = 1
Solution:
x^y + y^x = 1
let u = x^y and v = y^x
∴ u + v = 1,

Now u = x

Question 13.
y^x = x^y
Solution:
y = x
x logy = y logx

Question 14.
(cos x)^y = (cos y)^x
Solution:
We have
(cos x)^y = (cos y)^x
=> y log (cosx) = x log (cosy)

Question 15.
xy = e^(x-y)
Solution:
log(xy) = log e^(x-y)
=> log(xy) = x – y
=> logx + logy = x – y

Question 16.
Find the derivative of the function given by f (x) = (1 + x) (1 + x²) (1 + x⁴) (1 + x⁸) and hence find f'(1).
Solution:
Let f(x) = y = (1 + x)(1 + x²)(1 + x⁴)(1 + x⁸)
Taking log both sides, we get
logy = log [(1 + x)(1 + x²)(1 + x⁴)(1 + x⁸)]
logy = log(1 + x) + log (1 + x²) + log(1 + x⁴) + log(1 + x⁸)

Question 17.
Differentiate (x² – 5x + 8) (x³ + 7x + 9) in three ways mentioned below:
(i) by using product rule
(ii) by expanding the product to obtain a single polynomial.
(iii) by logarithmic differentiation.
Do they all give the same answer?
Solution:
(i) By using product rule
f’ = (x² – 5x + 8) (3x² + 7) + (x³ + 7x + 9) (2x – 5)
f = 5x⁴ – 20x³ + 45x² – 52x + 11.
(ii) By expanding the product to obtain a single polynomial, we get

Question 18.
If u, v and w are functions of w then show that

in two ways-first by repeated application of product rule, second by logarithmic differentiation.
Solution:
Let y = u.v.w
=> y = u. (vw)

 

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