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

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Here, Below you all know about NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Question Answer. I know many of you confuse about finding Chapter 5 Continuity and Differentiability Ex 5.5 Of Class 12 NCERT Solutions. So, Read the full post below and get your solutions.

TextbookNCERT
BoardCBSE
CategoryNCERT Solutions
ClassClass 12
SubjectMaths
ChapterChapter 5
ExerciseClass 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.5
Number of Questions Solved18
NCERT Solutions For Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5

Table of Contents

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

NCERT TEXTBOOK EXERCISES

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

Ex 5.5 Class 12 Maths Question 2.
$\sqrt { \frac { (x-1)(x-2) }{ (x-3)(x-4)(x-5) } }$

Solution:

y= $\sqrt { \frac { (x-1)(x-2) }{ (x-3)(x-4)(x-5) } }$
taking log on both sides
log y = log $\sqrt { \frac { (x-1)(x-2) }{ (x-3)(x-4)(x-5) } }$

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,

Ex 5.5 Class 12 Maths Question 4.
x – 2sinx

Solution:

let y = x – 2sinx,
y = u – v

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

Ex 5.5 Class 12 Maths Question 6.
${ \left( x+\frac { 1 }{ x } \right) }^{ x }+{ x }^{ \left( 1+\frac { 1 }{ x } \right) }$

Solution:

let y= ${ \left( x+\frac { 1 }{ x } \right) }^{ x }+{ x }^{ \left( 1+\frac { 1 }{ x } \right) }$
let $u={ \left( x+\frac { 1 }{ x } \right) }^{ x }and\quad v={ x }^{ \left( 1+\frac { 1 }{ x } \right) }$

Ex 5.5 Class 12 Maths Question 7.
(log x)x + xlogx

Solution:

let y = (log x)x + xlogx = u+v
where u = (log x)x
∴ log u = x log(log x)

Ex 5.5 Class 12 Maths Question 8.
(sin x)x+sin-1 √x

Solution:

Let y = (sin x)+ sin-1 √x
let u = (sin x)x, v = sin-1 √x

Ex 5.5 Class 12 Maths Question 9.
xsinx + (sin x)cosx

Solution:

let y = xsinx + (sin x)cosx = u+v
where u = xsinx
log u = sin x log x

Ex 5.5 Class 12 Maths Question 10.
${ x }^{ x\quad cosx }+\frac { { x }^{ 2 }+1 }{ { x }^{ 2 }-1 }$

Solution:

y= ${ x }^{ x\quad cosx }+\frac { { x }^{ 2 }+1 }{ { x }^{ 2 }-1 }$
y = u + v

Ex 5.5 Class 12 Maths Question 11.
${ (x\quad cosx) }^{ x }+{ (x\quad sinx) }^{ \frac { 1 }{ x } }$

Solution:

y= ${ (x\quad cosx) }^{ x }+{ (x\quad sinx) }^{ \frac { 1 }{ x } }$
Let u = (x cosx)x
logu = x log(x cosx)

Find dy/dx of the functions given in Questions 12 to 15.

Ex 5.5 Class 12 Maths Question 12.
xy + yx = 1

Solution:

xy + yx = 1
let u = xy and v = yx
∴ u + v = 1,
$\frac { du }{ dx } +\frac { dv }{ dx }=0$
Now u = x

Ex 5.5 Class 12 Maths Question 13.
y= xy

Solution:

y = x
x logy = y logx

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)

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
=> $=>\frac { 1 }{ x } +\frac { 1 }{ y } \frac { dy }{ dx } =1-\frac { dy }{ dx } =>\frac { dy }{ dx } =\frac { y(x-1) }{ x(y+1) }$

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)

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

Ex 5.5 Class 12 Maths Question 18.
If u, v and w are functions of w then show that
$\frac { d }{ dx } (u.v.w)=\frac { du }{ dx } v.w+u.\frac { dv }{ dx } .w+u.v\frac { dw }{ dx }$
in two ways-first by repeated application of product rule, second by logarithmic differentiation.

Solution:

Let y = u.v.w
=> y = u. (vw)

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

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