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 5.4

Get Free NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Continuity and Differentiability Exercise 5.5 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 5 Class 12 Continuity and Differentiability Ex 5.5 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

The topics and sub-topics included in the Continuity and Differentiability chapter are the following:

Continuity and Differentiability
Introduction
Algebra of continuous functions
Differentiability
Derivatives of composite functions
Derivatives of implicit functions
Derivatives of inverse trigonometric functions
Exponential and Logarithmic Functions
Logarithmic Differentiation
Derivatives of Functions in Parametric Forms
Second Order Derivative
Mean Value Theorem
Summary

There are total eight exercises and one misc exercise(144 Questions fully solved) in the class 12th maths chapter 5 Continuity and Differentiability.

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

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

Ex 5.5 Class 12 Maths 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
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Ex 5.5 Class 12 Maths Question 2.
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Clik here to view. $\sqrt { \frac { (x-1)(x-2) }{ (x-3)(x-4)(x-5) } }$
Solution:
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Clik here to view. $y=\sqrt { \frac { (x-1)(x-2) }{ (x-3)(x-4)(x-5) } }$
taking log on both sides
log y = log Image may be NSFW.
Clik here to view. $\sqrt { \frac { (x-1)(x-2) }{ (x-3)(x-4)(x-5) } }$
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Ex 5.5 Class 12 Maths 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,
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Ex 5.5 Class 12 Maths Question 4.
x – 2^sinx
Solution:
let y = x – 2^sinx,
y = u – v
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Ex 5.5 Class 12 Maths 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
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Ex 5.5 Class 12 Maths Question 6.
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Clik here to view. ${ \left( x+\frac { 1 }{ x } \right) }^{ x }+{ x }^{ \left( 1+\frac { 1 }{ x } \right) }$
Solution:
let Image may be NSFW.
Clik here to view. $y={ \left( x+\frac { 1 }{ x } \right) }^{ x }+{ x }^{ \left( 1+\frac { 1 }{ x } \right) }$
let Image may be NSFW.
Clik here to view. $u={ \left( x+\frac { 1 }{ x } \right) }^{ x }and\quad v={ x }^{ \left( 1+\frac { 1 }{ x } \right) }$
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Ex 5.5 Class 12 Maths 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)
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Ex 5.5 Class 12 Maths 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
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Ex 5.5 Class 12 Maths 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
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Ex 5.5 Class 12 Maths Question 10.
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Clik here to view. ${ x }^{ x\quad cosx }+\frac { { x }^{ 2 }+1 }{ { x }^{ 2 }-1 }$
Solution:
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Clik here to view. $y={ x }^{ x\quad cosx }+\frac { { x }^{ 2 }+1 }{ { x }^{ 2 }-1 }$
y = u + v
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Ex 5.5 Class 12 Maths Question 11.
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Clik here to view. ${ (x\quad cosx) }^{ x }+{ (x\quad sinx) }^{ \frac { 1 }{ x } }$
Solution:
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Clik here to view. $y={ (x\quad cosx) }^{ x }+{ (x\quad sinx) }^{ \frac { 1 }{ x } }$
Let u = (x cosx)^x
logu = x log(x cosx)
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Clik here to view. NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 11
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Find Image may be NSFW.
Clik here to view. $\\ \frac { dy }{ dx }$ of the functions given in Questions 12 to 15.

Ex 5.5 Class 12 Maths Question 12.
x^y + y^x = 1
Solution:
x^y + y^x = 1
let u = x^y and v = y^x
∴ u + v = 1,
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Clik here to view. $\frac { du }{ dx } +\frac { dv }{ dx }=0$
Now u = x
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Ex 5.5 Class 12 Maths Question 13.
y^x= x^y
Solution:
y = x
x logy = y logx
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Ex 5.5 Class 12 Maths Question 14.
(cos x)^y = (cos y)^x
Solution:
We have
(cos x)^y = (cos y)^x
=> y log (cosx) = x log (cosy)
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Clik here to view. NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 14

Ex 5.5 Class 12 Maths Question 15.
xy = e^(x-y)
Solution:
log(xy) = log e^(x-y)
=> log(xy) = x – y
=> logx + logy = x – y
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Clik here to view. $=>\frac { 1 }{ x } +\frac { 1 }{ y } \frac { dy }{ dx } =1-\frac { dy }{ dx } =>\frac { dy }{ dx } =\frac { y(x-1) }{ x(y+1) }$

Ex 5.5 Class 12 Maths 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)
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Ex 5.5 Class 12 Maths 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
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Ex 5.5 Class 12 Maths Question 18.
If u, v and w are functions of w then show that
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Clik here to view. $\frac { d }{ dx } (u.v.w)=\frac { du }{ dx } v.w+u.\frac { dv }{ dx } .w+u.v\frac { dw }{ dx }$
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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NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Hindi Medium Ex 5.5

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Clik here to view. NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 in PDF
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Clik here to view. 12 Maths exercise 5.5 in English medium
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Clik here to view. 12th maths Exercise 5.5 Solutions
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Clik here to view. NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5 question 7, 8, 9, 10
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Clik here to view. class 12 maths ex. 5.5 question 11, 12, 13, 14, 15
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Clik here to view. 12 Maths exercise 5.5 in English medium
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Clik here to view. 12 Maths exercise 5.5 in Hindi medium
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Clik here to view. 12 Maths exercise 5.5 updated for 2018-19 CBSE
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