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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
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 5.1

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
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 9.1

 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
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 11.1

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
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 17.1

 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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