Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.4, Exercise 5.5, Exercise 5.6

Exercise 5.4

Differentiate the following w.r.t.x:

Question 1.
$\frac { { e }^{ x } }{ sinx }$
Solution:
$y=\frac { { e }^{ x } }{ sinx }$
$for\quad y=\frac { u }{ v } ,$
$\frac { dy }{ dx } =\frac { { e }^{ x }{ sin }x-{ e }^{ x }cosx }{ { sin }^{ 2 }x }$
$or\frac { dy }{ dx } =\frac { { e }^{ x }{ sin }x-{ e }^{ x }cosx }{ { sin }^{ 2 }x } ,where\quad x\neq n\pi ,x\in z$

Question 2.
${ e }^{ { sin }^{ -1 }x }$
Solution:
${ e }^{ { sin }^{ -1 }x }$
$y={ e }^{ { sin }^{ -1 }x }$
x=sint
$\therefore y={ e }^{ t },\frac { dt }{ dx } =\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } ,\frac { dy }{ dt } ={ e }^{ t }$
$\therefore \frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } ={ e }^{ t }.\frac { 1 }{ \sqrt { { 1- }x^{ 2 } } } =\frac { { e }^{ { sin }^{ -1 }x } }{ \sqrt { 1-{ x }^{ 2 } } }$

Question 3.
${ e }^{ { x }^{ 3 } }=y$
Solution:
${ e }^{ { x }^{ 3 } }=y$
$Put\quad { x }^{ 3 }=t\quad \therefore \quad y={ e }^{ t },\frac { dy }{ dt } ={ e }^{ t },\frac { dt }{ dx } ={ 3x }^{ 2 }$
$\therefore \frac { dy }{ dx } =\frac { dy }{ dt } \times \frac { dt }{ dx } ={ e }^{ t }\times { 3x }^{ 2 }={ 3x }^{ 2 }{ e }^{ { x }^{ 3 } }$

Question 4.
$sin\left( { tan }^{ -1 }{ e }^{ -x } \right) =y$
Solution:
$sin\left( { tan }^{ -1 }{ e }^{ -x } \right) =y$
$\frac { dy }{ dx } =cos\left( { tan }^{ -1 }{ e }^{ -x } \right) \frac { d }{ dx } \left( { tan }^{ -1 }{ e }^{ -x } \right)$
$=cos\left( { tan }^{ -1 }{ e }^{ -x } \right) \frac { 1 }{ 1+{ e }^{ -2x } } \frac { d }{ dx } \left( { e }^{ -x } \right)$
$=-cos\left( { tan }^{ -1 }{ e }^{ -x } \right) \frac { 1 }{ 1+{ e }^{ -2x } } .\left( { e }^{ -x } \right)$

Question 5.
$log(cos\quad { e }^{ x })=y$
Solution:
$\frac { dy }{ dx } =\frac { 1 }{ cos\quad { e }^{ x } } \left( -sin{ e }^{ x } \right) .{ e }^{ x }\quad =-tan\left( { e }^{ x } \right)$

Question 6.
${ e }^{ x }+{ e }^{ { x }^{ 2 } }+$ … $+{ e }^{ { x }^{ 5 } }=y(say)$
Solution:
$let\quad u={ e }^{ { x }^{ n } },put\quad { x }^{ n }=t,u={ e }^{ t },t={ x }^{ n }$
${ e }^{ x }+{ e }^{ { x }^{ 2 } }+$ … $+{ e }^{ { x }^{ 5 } }=y(say)$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 6

Question 7.
$\sqrt { { e }^{ \sqrt { x } } } ,x>0$
Solution:
y = $\sqrt { { e }^{ \sqrt { x } } } ,x>0$
$y=\sqrt { { e }^{ \sqrt { x } } } ,let\quad y=\sqrt { s } ,s={ e }^{ t },t=\sqrt { x }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 7

Question 8
log(log x),x>1
Solution:
y = log(log x),
put y = log t, t = log x,
differentiating
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 8

Question 9.
$\frac { cosx }{ logx } =y(say),x>0$
Solution:
let $y=\frac { cosx }{ logx }$
tiwari academy class 12 maths Chapter 5 Continuity and Differentiability 9

Question 10.
cos(log x+ex),x>0
Solution:
y = cos(log x+ex),x>0
put y = cos t,t = log x+ex
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 10

Exercise 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
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 1

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 $\sqrt { \frac { (x-1)(x-2) }{ (x-3)(x-4)(x-5) } }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 2

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,
vedantu class 12 maths Chapter 5 Continuity and Differentiability 3

Question 4.
x – 2sinx
Solution:
let y = x – 2sinx,
y = u – v
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 4

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

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 $u={ \left( x+\frac { 1 }{ x } \right) }^{ x }and\quad v={ x }^{ \left( 1+\frac { 1 }{ x } \right) }$
vedantu class 12 maths Chapter 5 Continuity and Differentiability 6
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 6.1

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)
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 7

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
vedantu class 12 maths Chapter 5 Continuity and Differentiability 8

Question 9.
xsinx + (sin x)cosx
Solution:
let y = xsinx + (sin x)cosx = u+v
where u = xsinx
log u = sin x log x
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 9

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
vedantu class 12 maths Chapter 5 Continuity and Differentiability 10

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)
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 11

Find $\\ \frac { dy }{ dx }$ of the functions given in Questions 12 to 15.

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
vedantu class 12 maths Chapter 5 Continuity and Differentiability 12

Question 13.
yx = xy
Solution:
y = x
x logy = y logx
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 13

Question 14.
(cos x)y = (cos y)x
Solution:
We have
(cos x)y = (cos y)x
=> y log (cosx) = x log (cosy)
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 14

Question 15.
xy = e(x-y)
Solution:
log(xy) = log e(x-y)
=> log(xy) = x – y
=> logx + logy = x – y
$=>\frac { 1 }{ x } +\frac { 1 }{ y } \frac { dy }{ dx } =1-\frac { dy }{ dx } =>\frac { dy }{ dx } =\frac { y(x-1) }{ x(y+1) }$

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)
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 16

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
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 17

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 18

Exercise 5.6

If x and y are connected parametrically by the equations given in Questions 1 to 10, without eliminating the parameter. Find $\\ \frac { dy }{ dx }$ .

Question 1.
x = 2at², y = at4
Solution:
$\frac { dx }{ dt } =4at,\frac { dy }{ dt } ={ 4at }^{ 3 }\quad \therefore \frac { dy }{ dx } =\frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } =\frac { { 4at }^{ 3 } }{ { 4at } } ={ t }^{ 2 }$

Question 2.
x = a cosθ,y = b cosθ
Solution:
$\frac { dx }{ d\theta } =-asin\theta ,\frac { dy }{ d\theta } =-sinb\quad sin\theta =>\frac { dy }{ dx } =\frac { b }{ a }$

Question 3.
x = sin t, y = cos 2t
Solution:
$\therefore \frac { dx }{ dt } =cos\quad t\quad and\frac { dy }{ dt } =-sin2t.2=-2sin2t$
$\frac { dy }{ dx } =\frac { -2sin2t }{ cost } =\frac { -2.2sintcost }{ cost } =-4sint$

Question 4.
$x=4t,y=\frac { 4 }{ t }$
Solution:
$\frac { dx }{ dt } =4;\frac { dy }{ dt } =\frac { -4 }{ { t }^{ 2 } } =>\frac { dy }{ dx } =\frac { -4 }{ { t }^{ 2 } } \times \frac { 1 }{ 4 } =\frac { -1 }{ { t }^{ 2 } }$

Question 5.
x = cos θ – cos 2θ, y = sin θ – sin 2θ
Solution:
$\frac { dx }{ d\theta } =-sin\theta -(-sin2\theta ).2=2sin2\theta -sin\theta$
$\frac { dy }{ d\theta } =cos\theta -2cos2\theta \quad \therefore \frac { dy }{ dx } =\frac { cos\theta -2cos2\theta }{ 2sin2\theta -sin\theta }$

Question 6.
x = a(θ – sinθ), y = a(1 + cosθ)
Solution:
$\frac { dx }{ d\theta } =a\left[ 1-cos\theta \right] \& \frac { dy }{ d\theta } =-asin\theta$
$\frac { dy }{ dx } =\frac { -asin\theta }{ a(1-cos\theta ) } =\frac { -2sin\frac { \theta }{ 2 } .cos\frac { \theta }{ 2 } }{ 2{ sin }^{ 2 }\frac { \theta }{ 2 } } =-cot\frac { \theta }{ 2 }$

Question 7.
$x=\frac { { sin }^{ 3 }t }{ \sqrt { cos2t } } \& y=\frac { { cos }^{ 3 }t }{ \sqrt { cos2t } }$
Solution:
$x=\frac { { sin }^{ 3 }t }{ \sqrt { cos2t } } \& y=\frac { { cos }^{ 3 }t }{ \sqrt { cos2t } }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 7

Question 8.
$x=a\left( cost+log\quad tan\frac { t }{ 2 } \right) ,y=a\quad sint$
Solution:
$x=a\left( cost+log\quad tan\frac { t }{ 2 } \right) ,y=a\quad sint$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 8

Question 9.
x = a sec θ,y = b tan θ
Solution:
x = a sec θ,y = b tan θ
$\frac { dx }{ d\theta } =a\quad sec\theta \quad tan\theta \quad and\frac { dy }{ d\theta } =b{ sec }^{ 2 }\theta$
$\frac { dy }{ dx } =\frac { { bsec }^{ 2 }\theta }{ asec\theta tan\theta } \frac { b }{ a } cosec\theta$

Question 10.
x = a(cosθ+θsinθ), y = a(sinθ-θcosθ)
Solution:
x = a(cosθ+θsinθ), y = a(sinθ-θcosθ)
$\frac { dx }{ d\theta } =a\left[ -sin\theta +\theta .cos\theta +sin\theta \right] =a\theta cos\theta$
$\frac { dy }{ d\theta } =a\theta sin\theta =>\frac { dy }{ dx } =\frac { a\theta sin\theta }{ a\theta cos\theta } =tan\theta$

Question 11.
If $x=\sqrt { { a }^{ { sin }^{ -1 }t } } ,y=\sqrt { { a }^{ { cos }^{ -1 }t } }$ show that $\frac { dy }{ dx } =-\frac { y }{ x }$
Solution:
Given that
$x=\sqrt { { a }^{ { sin }^{ -1 }t } } ,y=\sqrt { { a }^{ { cos }^{ -1 }t } }$
byjus class 12 maths Chapter 5 Continuity and Differentiability 11

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Verbs What is a verb? Verbs are the action words in a sentence that describe what the subject is doing. Along with nouns, verbs are the main part of a sentence or phrase, telling a story about what is taking place. Verbs are words that express action or state of being. There are three types of verbs: action verbs, linking verbs, and helping verbs . Action verbs are words that express action (give, eat, walk, etc.) or possession (have, own, etc.). Action verbs can be either transitive or intransitive. List of Verb in hindi Present Hindi Meaning Past Past Participle Buy खरीदना Bought Bought Build बांधना Built Built burn जलना Burnt Burnt Bend झुकना Bent Bent Bring लाना brought brought Become होना Became Become Come आना Came Come Catch पकड़ना Caught Caught Do करना Did Done Dream ख्वाब देखना Dreamt Dreamt Arise उठना / जागना Arose Arisen Be होना Was, were Been Bear सहन करना Bore Bore Beat मारना Beat Beat Bite काटना Bit Bitten Break तोडना Broke Broken Choose चुनना Chose Chosen Draw चि

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Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.4, Exercise 5.5, Exercise 5.6

Exercise 5.6

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