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RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise Differentiation Excercise: 10.3 RD Sharma Chapter-wise Solutions Also Read - RD Sharma Solution for Class 9 to 12 Maths
RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise Differentiation Excercise: 10.3 Answer :
d y d x = 2 1 − x 2 Hint:
cos − 1 ( cos θ ) = θ i f θ ε [ 0 , π ] cos − 1 ( cos θ ) = − θ i f θ ε [ − π , 0 ] Given:
cos − 1 { 2 x 1 − x 2 } , 1 2 < x < 1 Solution:
L e t s y = cos − 1 { 2 x 1 − x 2 } L e t s x = cos θ ⇒ θ = cos − 1 x N o w y = cos − 1 { 2 cos θ 1 − cos 2 θ } U s i n g ∴ cos 2 θ + sin 2 θ = 1 , 2 sin θ cos θ = sin 2 θ y = cos − 1 { 2 cos θ sin 2 θ } y = cos − 1 ( 2 cos θ sin θ ] y = cos − 1 ( sin 2 θ ) ∴ cos ( π 2 − θ ) ⇒ sin θ y = cos − 1 ( cos ( π 2 − 2 θ ) ) Now by considering the limits,
1 2 < x < 1 ⇒ 1 2 < cos θ < 1 ⇒ 0 < θ < π 4 ∴ [ cos 90 ∘ = 0 , cos 0 ∘ = 1 ] ⇒ 0 < 2 θ < π 2 ⇒ 0 > − 2 θ > − π 2 ⇒ π 2 > π 2 − 2 θ > π 2 − π 2 Therefore,
y = cos − 1 ( cos ( π 2 − 2 θ ) ) y = cos − 1 ( cos ( π 2 − 2 θ ) ) y = ( π 2 − 2 θ ) { cos − 1 ( cos θ ) = θ } y = π 2 − 2 cos − 1 x if θ ε [ 0 , π ] Differentiating with Respect to x, we get
d y d x = d d x ( π 2 − 2 cos − 1 x ) ∴ d d x ( cos − 1 x ) ⇒ − 1 1 − x 2 ∴ d dx (constant) = 0 d y d x = 0 − 2 ( − 1 1 − x 2 ) d y d x = 2 1 − x 2 Differentiation exercise 10.3 question 2
Answer :
dy dx = 1 2 ( − 1 1 − x 2 ) Hint:
cos − 1 ( cos θ ) = θ i f θ ε [ 0 , π ] cos − 1 ( cos θ ) = − θ i f θ ε [ − π , 0 ] Given:
cos − 1 { 1 + x 2 } , − 1 < x < 1 Solution:
L e t , y = cos − 1 { 1 + x 2 } L e t x = cos 2 θ , θ = 1 2 cos − 1 x N o w , y = cos − 1 { 1 + cos 2 θ 2 } B y u s i n g cos 2 θ = 2 cos 2 θ − 1 y = cos − 1 { 2 cos 2 θ 2 } y = cos − 1 ( cos θ ) ∴ cos − 1 ( cos θ ) = θ N o w y = cos − 1 ( cos θ ) y = θ y = 1 2 cos − 1 x ∴ cos − 1 ( cos θ ) = θ i f θ ε [ 0 , π ] Differentiating with respect to x, we get
dy dx ⇒ 1 2 ( − 1 1 − x 2 ) As we know,
d dx ( c o n s t a n t s ) = 0 d d x ( cos − 1 x ) = − 1 1 − x 2 Hence, final answer is
1 2 ( − 1 1 − x 2 ) Differentiation exercise 10.3 question 3
Answer : -
− 1 2 1 − x 2 Hint:
Use substitution method to differentiate this function
Given:
sin − 1 { 1 − x 2 } } , 0 < x < 1 Solution:
Let y = sin − 1 { 1 − x 2 } let x = cos 2 θ [ θ = 1 2 cos − 1 x ] − ( 1 ) ∴ y = sin − 1 { 1 − cos 2 θ 2 } = sin − 1 { 2 sin 2 θ 2 } y = sin − 1 { sin 2 θ } [ ∵ 1 − cos 2 θ = 2 sin 2 θ ] = sin − 1 ( sin θ ) y = θ = 1 2 cos − 1 x [ using (i)] Now differentiating y w.r.t x then
d y d x = 1 2 d ( cos − 1 x ) d x = 1 2 ⋅ − 1 1 − x 2 = − 1 2 1 − x 2 [ ∵ d d x cos − 1 x = − 1 1 − x 2 ] ∴ d y d x = − 1 2 1 − x 2 Differentiation exercise 10.3 question 4
Answer :
d y d x = − 1 1 − x 2 Hint:
d d x ( x n ) = n x n − 1 ; d d x ( c o n s t a n t ) = 0 Given:
sin − 1 { 1 − x 2 } , 0 < x < 1 Solution:
Unknown environment '' Unknown environment '' Considering the limits,
Unknown environment '' Unknown environment '' differentiating with respects to x, we get
Unknown environment '' Unknown environment '' Differentiation exercise 10.3 question 6
Answer :
d y d x = a a 2 + x 2 Hint:
Given:
sin − 1 { x x 2 + a 2 } Solution:
L e t y = sin − 1 { − x x 2 + a 2 } L e t x = atan θ N o w , y = sin − 1 { atan θ a 2 tan 2 θ + a 2 } y = sin − 1 { atsan θ a 2 ( tan 2 θ + 1 ) } y = sin − 1 { atan θ a ( tan 2 θ + 1 ) } y = sin − 1 ∫ atan θ a sec 2 θ } Using : { tan θ = sin θ cos θ , cos θ = 1 sec θ } y = sin − 1 { sin θ } y = θ ∴ { sin − 1 θ ( sin θ ) = θ } if θ ε [ − π 2 , π 2 ] x = atan θ x = atan θ θ = tan − 1 ( x a ) y = tan − 1 ( x a ) differentiating with respects to x, we get
d y d x = d d x ( tan − 1 ( x a ) ) ∂ ∂ x ( tan − 1 x ) = 1 1 + x 2 d y d x = ( 1 1 + ( x a ) 2 ) × 1 a d y d x = ( 1 1 + x 2 a 2 ) × 1 a d y d x = ( 1 a 2 + x 2 a 2 ) × 1 a d y d x = a 2 a 2 + x 2 × 1 a d y d x = a a 2 + x 2 Differentiation exercise 10.3 question 7
Answer:
d y d x = 2 1 − x 2 Hint:
d dx ( C o n s t a n t ) = 0 ; ∂ dx ( x n ) = n x n − 1 Given:
sin − 1 ( 2 x 2 − 1 ) , 0 < x < 1 Solution:
Unknown environment '' Unknown environment '' As we know,
sin ( n 2 − θ ) = cos θ Considering the limits,
0 < x < 1 Unknown environment '' Unknown environment '' Differentiating with respect ts? x, weget
d y d x = d ∂ x ( n 2 − 2 cos − 1 x ) As we know
Unknown environment '' Unknown environment '' Differentiation exercise 10.3 question 8 : dy dx = − 2 1 − x 2
Hint:
∂ d x ( C o n s t a n t ) = 0 ; ∂ dx ( x n ) = n x n − 1 Given:
sin − 1 ( 1 − 2 x 2 ) , 0 < x < 1 Solution:
Unknown environment '' Unknown environment '' As we know,
Unknown environment '' Unknown environment '' Considering the limits,
Unknown environment '' Unknown environment '' Now,
Unknown environment '' Unknown environment '' Differentiating with respect to x,
d y d x = d d x ( n 2 − 2 sin − 1 x ) As we know d d x ( constant ) = 0 d d x ( sin − 1 x ) = 1 1 − x 2 d y d x = 0 − 2 1 1 − x 2 d y d x = − 2 1 − x 2 Differentiation exercise 10.3 question 9
Answer :
d y d x = − a a 2 + x 2 Hint:
d d x ( C o n s t a n t ) = 0 d d x ( x n ) = n x n − 1 Given:
cos − 1 { x x 2 + a 2 } Solution:
L e t , y = cos − 1 { x x 2 + a 2 ) L e t x = acot θ θ = cot − 1 ( x a ) Now y = cos − 1 { acot θ a 2 cot 2 θ + a 2 } y = cos − 1 { acot θ a 2 ( cot 2 θ + 1 ) } y = cos − 1 { acot θ a cot 2 θ + 1 } Using 1 + cot 2 θ = cosec 2 θ y = cos − 1 { acot θ a cosec 2 θ } y = cos − 1 { acot θ acosec θ } y = cos − 1 { cot θ cosec θ } As we Know,
cot θ = cos θ sin θ , cosec θ = 1 sin θ y = cos − 1 { cos θ sin θ × sin θ } y = cos − 1 ( cos θ ) y = θ y = cot − 1 ( x a ) Differentiating with respect to x We get d y d x = d d x ( cot − 1 ( x a ) ) ∴ ∂ d x ( cot − 1 x ) = − 1 1 + x 2 d y dx = − 1 1 + ( x a ) 2 × 1 a d y d x = − 1 1 + x 2 a 2 × 1 a d y d x = − 1 a 2 + x 2 a 2 × 1 a d y d x = a 2 a 2 + x 2 × 1 a d y d x = − a a 2 + x 2 Differentiation Exercise 10.3 Question 10
Answer:
d y d x = 1 Hint:
d d x (constants) = 0 d d x ( x n ) = nx n − 1 Given:
sin − 1 { sin x + cos x 2 } − 3 π 4 < x < π 4 Soluton:
Let,
y = sin − 1 { sin x + cos x 2 } Now,
y = sin − 1 { sin x 1 2 + cos x 1 2 } y = sin − 1 { sin x cos ( π 4 ) + cos xsin ( π 4 ) } sin ( π 4 ) = 1 2 & cos ( π 4 ) = 1 2 Using,
sin ( A + B ) = sin A cos B + cos A sin B y = sin − 1 { sin ( x + π 4 ) } Considering the limits, − 3 π 4 < x < π 4 − 3 π 4 + π 4 < x + π 4 < π 4 + 1 4 − π 2 < x + π 4 < π 2 Now, y = sin − 1 { sin ( x + π 4 ) } y = x + π 4 { sin − 1 ( sin θ ) = θ , if θ ε [ π 2 , π 2 ] } Differentiating with respect to x , We get
d y d x = 1 + 0 d dx (constant) = 0 d y d x = 1 Differentiation Exercise 10.3 Question 11
Answer:
d y d x = − 1 Hint:
d dx (constants) = 0 d d x ( x n ) = n x n − 1 Given:
cos − 1 { cos x + sin x 2 } − π 4 < x < π 4 Solution:
Let,
y = cos − 1 { cos x + sin x 2 } Now
y = cos − 1 { cos x 1 2 + sin x 1 2 } y = cos − 1 { cos x cos ( π 4 ) + sin x sin ( π 4 ) } sin ( π 4 ) = 1 2 & cos ( π 4 ) = 1 2 Using,
cos ( A − B ) = cos A cos B + sin A sin B y = cos − 1 { cos ( x − π 4 ) } Considering the limits,
− π 4 < x < π 4 − π 4 − π 4 < x − π 4 < π 4 − π 4 − 2 π 4 < x − π 4 < 0 − π 2 < x − π 4 < 0 Now,
y = cos − 1 { cos ( x − π 4 ) } y = − ( x − π 4 ) { cos − 1 ( cos θ ) = − θ if θ ∈ [ − π , 0 ] } Differentiating with respect to x , We get
d y d x = − 1
d d x ( constants ) = 0
Differentiation Exercise 10.3 Question 12 (i)
Answer:
d y d x = 1 2 1 − x 2 Hint:
d d x ( c o n s t a n t ) = 0 d d x ( x n ) = n x n − 1 Given:
tan − 1 { x 1 + 1 − x 2 } − 1 < x < 1 Solution:
Let,
y = tan − 1 { x 1 + 1 − x 2 } x = sin θ θ = sin − 1 x Now,
y = tan − 1 { sin θ 1 + 1 sin 2 θ } Using
sin 2 θ + cos 2 θ = 1 y = tan − 1 { sin θ 1 + cos 2 θ } y = tran − 1 { sin θ 1 + cos θ } Using 2 cos 2 θ = 1 + cos 2 θ 2 sin θ cos θ = sin 2 θ y = tan − 1 { 2 sin θ 2 cos θ 2 2 cos 2 θ 2 } y = tan − 1 { sin θ 2 cos θ 2 } tan θ = sin θ cos θ y = tan − 1 { tan θ 2 } Considering the limits,
− 1 < x < 1 − 1 < sin θ < 1 − π 2 < θ 2 < π 2 − π 4 < θ 2 < π 4
Now,
y = tan − 1 { tan θ 2 } y = θ 2 tan − 1 ( tan θ ) = 0 if θ ∈ [ − π 2 , π 2 ]
y = 1 2 sin − 1 x
Differentiating with respect to x , We get
d y d x = − 1 d d x ( constants ) = 0 d y d x = 1 2 ( 1 2 d d x ( sin − 1 x ) d d x ( sin − 1 x ) = 1 1 − x 2 d y d x = 1 2 1 1 − x 2 d y d x = 1 2 1 − x 2
− 1 < x < 1
Differentiation Exercise 10.3 Question 12 (ii)
Answer:
d y d x = 1 2 Hint:
cos 2 x = 2 cos 2 x − 1 tan ( π 2 − θ ) = cot θ Given:
tan − 1 { 1 + cos x s i n x } Solution:
y = tan − 1 { 1 + cos x sin x } y = tan − 1 { 2 cos 2 x 2 2 sin x 2 cos x 2 } y = tan − 1 ( cot x 2 ) y = tan − 1 ( tan ( π 2 − x 2 ) ) [ ( π 2 − θ ) = cot θ ] y = x 2 Differentiating w.r.t x d y d x = 1 2 Differentiation Exercise 10.3 Question 13
Answer:
d y d x = 1 2 a 2 − x 2 Hint:
d dx ( constants ) = 0 d dx ( x n ) = n x n − 1 Given:
tan − 1 { x a + a 2 − x 2 } − a < x < − a Solution:
Let,
x = a sin θ θ = sin − 1 x a Now,
y = tan − 1 { asin θ a + a 2 − a 2 sin 2 θ } y = tan − 1 { asin θ a + a 1 − sin 2 θ } Using
cos 2 θ + sin 2 θ = 1 y = tan − 1 { asin θ a + a cos θ } y = tan − 1 { asin θ a + acos θ } y = tan − 1 { asin θ a ( 1 + los θ ) } U sing 2 cos 2 θ = 1 + cos θ y = tan − 1 { sin θ 1 + cos θ } and 2 sin θ cos θ = sin 2 θ y = tan − 1 { sin θ 1 + cos θ } y = tan − 1 { 2 sin θ 2 cos θ 2 2 cos 2 θ 2 } y = tan − 1 { sin θ 2 cos θ 2 } y = tan − 1 ( tan θ 2 ) ∴ sin θ cos θ = tan θ Considering the limit − a < x < a − 1 < sin θ < 1 − π 2 < θ < π 2 − π 4 < θ 2 < π 4 { sin π 2 = 1 } Now,
y == tan − 1 { tan θ 2 } y = θ 2 y = 1 2 sin − 1 x a Differentiating with respect to x , We get
d y d x = d d x ( 1 2 sin − 1 x a ) ∴ d d x ( sin − 1 x ) = 1 1 − x 2 d y d x = 1 2 1 1 − ( x ) 2 × 1 a ⇒ 1 2 × 1 1 − x 2 a 2 × 1 a ⇒ d y d x = 1 2 × 1 a 2 − x 2 a 2 × 1 a d y d x = 1 2 a 2 − x 2
Differentiation Exercise 10.3 Question 14
Answer:
d y d x = 1 1 − x 2 Hint:
d dx ( constants ) = 0 d d x ( x n ) = nx n − 1 Given:
sin − 1 { x + 1 − x 2 2 } − 1 < x < 1 Solution:
Let
y = sin − 1 { x + 1 − x 2 2 } Let
x = sin θ θ = sin − 1 x Now,
y = sin − 1 { sin θ + 1 − sin 2 θ 2 } Using
sin 2 θ + cos 2 θ = 1 y = sin − 1 { sin θ + cos 2 θ 2 ] y = sin − 1 { sin θ + cos θ 2 } Now,
y = sin − 1 { sin θ 1 2 + cos θ 1 2 } y = sin − 1 { sin θ cos ( π 4 ) + cos θ sin ( π 4 ) } sin ( π 4 ) = 1 2 & cos ( π 4 ) y = sin − 1 { sin ( θ + π 4 ) } Considering the limits,
− 1 < x < 1 − 1 < sin θ < 1 − π 2 < θ < π 2 { sin π 2 = 1 }
− π 2 + π 4 < θ + π 4 < π 2 + π 4 − π 4 < θ + π 4 < 3 π 4
Now,
y = sin − 1 { sin ( θ + π 4 ) } y = θ + π 4 y = sin − 1 x + π 4 s i n − ( sin θ ) = θ i f θ ε [ − π 2 , π 2 ] d dx ( sin − 1 x ) = 1 1 − x 2 d d x (constant) = 0 dy dx = 1 1 − x 2
Differentiation Exercise 10.3 Question 15
Answer:
d y d x = − 1 1 − x 2 Hint:
d dx ( constants ) = 0 d d x ( x n ) = n x n − 1 Given:
cos − 1 { x + 1 − x 2 2 } − 1 < x < 1 Solution:
y = cos − 1 { x + 1 − x 2 2 } Let,
x = sin θ θ = sin − 1 x y = cos − 1 { sin θ + 1 − sin 2 θ 2 } y = cos − 1 { sin θ + cos 2 θ 2 } Using sin 2 θ + cos 2 θ = 1 y = cos − 1 { sin θ + cos θ 2 } y = cos − 1 { sin θ 1 2 + cos θ 1 2 } y = cos − 1 { sin θ sin ( π 4 ) + + cos θ cos ( π 4 ) } ∴ sin ( π 4 ) = 1 2 cos ( π 4 ) = 1 2 Using,
cos ( A − B ) = cos A cos B + sin A sin B y = cos − 1 { cos ( θ − π 4 ) } Considering the limits, − 1 < x < 1 − 1 < sin θ < 1 − π 2 < θ < π 2 − π 2 − π 4 < θ − π 4 < 1 2 − π 4 − 3 π 4 < 8 − π 4 < π 4 Now, y = cos − 1 { cos ( θ − π 4 ) } y = − ( θ − π 4 ) { ∴ cos − 1 ( cos θ ) = − 0 if θ ε [ − π , 0 ] } y = − ( θ − π 4 ) y = − sin − 1 x + π 4 [ s i n c e x = sin θ ] Differentiating with respect to x , We get d y d x = d d x ( − sin − 1 x + π 4 ) ∴ d d x ( constants ) = 0 d d x ( sin − 1 x ) = 1 1 − x 2 d y d x = − 1 1 − x 2 + 0 d y d x = − 1 1 − x 2 Differentiation Exercise 10.3 Question 16
Answer:
d y d x = 4 1 + 4 x 2 Hint:
d d x ( constants ) = 0 d d x ( x n ) = n x n − 1 Given:
tan − 1 { 4 x 1 − 4 x 2 } − 1 2 < x < 1 2 Solution:
Let,
y = tan − 1 { 4 x 1 − 4 x 2 } Let,
2 x = tan θ θ = tan − 1 2 x y = tan − 1 { 2 tan θ 1 − tan 2 θ } Using
tan 2 θ = 2 tan θ 1 − tan 2 θ y = tan − 1 ( tan 2 θ ) Considering the limits,
− 1 2 < x < 1 2 − 1 < 2 x < 1 − 1 < tan θ < 1 − π 4 < θ < π 4 − π 2 < 2 θ < π 2 { tan 45 ∘ = 1 }
Now,
y = tan − 1 ( tan 2 θ )
y = 2 θ
y = 2 tan − 1 ( 2 x ) [ s i n c e , 2 x = tan θ ]
Differentiating with respect to x , We get
d y d x = d d x ( 2 tan − 1 2 x ) d y d x = 2 × 2 1 + ( 2 x ) 2 d y d x = 4 1 + 4 x 2 d ( t a n − 1 x ) d x = 1 1 + x 2
Diffrentiation Exercise 10.3 Question 17
Answer:
d y d x = 2 x + 1 l o g 2 1 + 4 x Hint:
d dx ( constants ) = 0 d dx ( x n ) = n x n − 1 Given:
tan − 1 [ 2 x + 1 1 − 4 x ] − ∞ < x < 0 Solution:
y = tan − 1 [ 2 x + 1 1 − 4 x ] Let,
2 x = tan θ y = tan − 1 { 2 x 2 x 1 − ( 2 x ) 2 } As we Know
a m + n → a m ⋅ a n y = tan − 1 { 2 tan θ 1 − tan 2 θ } Using
tan 2 θ = 2 tan θ 1 − tan 2 θ y = tan − 1 ( tan 2 θ ) Considering the limits,
− ∞ < x < 0 2 − ∞ < 2 x < 2 0 0 < tan θ < 1 0 < π 4 { 2 0 = 1 , 2 − ∞ = 0 }
0 < 2 θ < π 2 { tan π 4 = 1 }
0 < 2 θ < π 2
Now
y = tan − 1 ( tan θ )
y == 2 θ
y = 2 tan − 1 ( 2 x ) { sin c e 2 x = tan θ θ = tan − 1 ( 2 x ) }
Differentiating with respect to x , We get
d y d x = d d x ( 2 tan − 1 ( 2 x ) ) d y d x = 2 × 2 x log 2 1 + ( 2 x ) 2 d y d x = 2 x + 1 log 2 1 + 4 x ∴ d ( tan − 1 x ) dx = 1 1 + x 2 d dx ( 2 x ) = log 2 { a m ⋅ a n ⇒ a m + h }
Differentiation Exercise 10.3 Question 18
Answer:
d y d x = 2 a x l o g a 1 + a 2 x Hint:
d c d x = 0 ; d d x ( x n ) = n x n − 1 Given:
tan − 1 { 2 a x 1 − a 2 x } , a > 1 , − ∞ < x < 0 Solution:
Let
y = tan − 1 { 2 a x 1 − a 2 x } , Let,
a x = tan θ y = tan − 1 { 2 tan θ 1 − tan 2 θ } Using tan 2 θ = 2 tan θ 1 − tan 2 θ y = tan − 1 ( tan 2 θ ) Considering the limits, − ∞ < x < 0 a − ∞ < a x < a 0 0 < tan θ < 1 { a o = 1 } 0 < θ < π 4 { tan ( π 4 ) = 1 } Now,
y = tan − 1 ( tan 2 θ ) y = 2 θ y = 2 tan − 1 ( a x ) s i n c e a x = tan θ , θ = tan − 1 ( a x ) Differentiating with respect to x we get d y d x = d d x ( 2 tan − 1 ( a x ) ) d y d x = 2 × a x 1 + ( a x ) 2 log a d d x ( tan − 1 x ) = 1 1 + x 2 d y d x = 2 a x log a 1 + a 2 x Differentiation Exercise 10.3 Question 19
Answer:
d y d x = − 1 2 1 − x 2 Hint:
d dx ( constants ) = 0 d dx ( x n ) = nx n − 1 Given:
sin − 1 { 1 + x + 1 − x 2 } , 0 < x < 1 Solution:
Let
y = sin − 1 { 1 + x + 1 − x 2 } Let,
x = cos 2 θ Now, sin − 1 { 1 + cos 2 θ + 1 − cos 2 θ 2 } Using 1 − 2 sin 2 θ = cos 2 θ 2 cos 2 θ − 1 = cos 2 θ y = sin − 1 { 2 cos 2 θ + 2 sin 2 θ 2 } Now,
y = sin − 1 { 2 cos θ + 2 sin θ 2 } y = sin − 1 { cos θ 1 2 + sin θ 1 2 } y = sin − 1 { sin θ cos ( π 4 ) + cos θ sin ( π 4 ) } ∴ sin ( π 4 ) = 1 2 & cos ( π 4 ) = 1 2 U sin sin ( A + B ) = sin A cos B + cos Asin B y = sin − 1 { sin ( θ + π 4 ) } Considering the limits, 0 < x < 1 0 < cos 2 θ < 1 0 < 2 θ < π 2 0 < θ < π 4 0 + π 4 < ( θ + 1 4 ) < π 2 + π 4 π 4 < ( θ + π 4 ) < π 2 Now,y = sin − 1 { sin ( θ + π 4 ) } y = θ + π 4 sin − 1 ( sin θ ) = θ , i f θ ε [ − π 2 , π 2 ] y = 1 2 cos − 1 x + π 4 { s i n c e x = cos 2 θ } d y d x = d d x ( 1 2 cos − 1 x + π 4 ) d dx ( constant ) = 0 d dx ( cos − 1 x ) = − 1 1 − x 2 dy dx = 1 2 ( − 1 1 − x 2 ) + 0 dy dx ⇒ − 1 2 1 − x 2 Differentiation Exercise 10.3 Question 20
Answer:
d y d x = a 2 ( 1 + a 2 x 2 ) Hint:
d d x ( constant ) = 0 d d x ( x n ) = n x n − 1 Given:
tan − 1 { 1 + a 2 x 2 − 1 a x } , x ≠ 0 Solution:
Let,
y = tan − 1 { 1 + a 2 x 2 − 1 a x } L e t a x = tan θ Now
y = tan − 1 { 1 + tan 2 θ − 1 tan θ } Using
sec 2 θ = 1 + tan 2 θ y = tan − 1 { sec 2 θ − 1 tan θ } y = tan − 1 { sec θ − 1 tan θ } y = tan − 1 { 1 cos θ − 1 sin θ cos θ } Using tan θ = sin θ cos θ , sec θ = 1 cos θ y = tan − 1 { 1 − cos θ cos θ sin d cos θ } y = tan − 1 { 1 − cos θ cos θ × cos θ sin θ } y = tan − 1 { 1 − cos θ sin θ } Using 2 sin 2 θ = 1 − cos 2 θ and 2 sin θ cos θ = sin 2 θ y = tan − 1 { 2 sin 2 θ / 2 2 sin θ / 2 cos θ / 2 } y = tan − 1 { tan θ 2 } y = θ 2 y = 1 2 t a n − 1 a x Differentiating with respect to x , We get d y d x = d d x ( 1 2 tan − 1 x ) Using d d x ( tan − 1 x ) ⇒ 1 1 + x 2 d y d x = 1 2 × 1 1 + ( a x ) 2 × 1 a d y d x = a 2 ( 1 + a 2 x 2 ) Differentiation Exercise 10.3 Question 21
Answer:
d y d x = 1 2 Hint:
d y d x ( x n ) = n x n − 1 d y d x ( c o n s t a n t ) = 0 Given:
tan − 1 { sin x 1 + cos x } , − π < x < π Solution:
Let,
y = tan − 1 { sin x 1 + cos x } Function y is defined for all the real numbers where cos x ≠ − 1 Using 2 cos 2 θ = 1 + cos 2 θ 2 sin θ cos θ = sin 2 θ y = tan − 1 { 2 sin x 2 cos x 2 2 cos 2 x 2 } y = tan − 1 { − sin x / 2 cos x / 2 } y = tan − 1 { tan x 2 } y = x 2 Differentiating with respect to x , We get d y d x = d d x ( x 2 ) d y d x = 1 2 Differentiation Exercise 10.3 Question 22
Answer:
d y d x = − 1 1 + x 2 Hint:
d d x ( Constant ) = 0 d d x ( x n ) = n x n − 1 Given:
sin − 1 { 1 1 + x 2 } Solution:
Let,
y = sin − 1 { 1 1 + x 2 } Let
x = cot θ θ = cot − 1 x Now,
y = sin − 1 { 1 1 + cot 2 θ } Using,
1 + cot 2 θ = cosec 2 θ y = sin − 1 { 1 cosec 2 θ } y = sin − 1 { 1 cosec θ } { 1 c o s e c θ = sin θ } y = sin − 1 ( sin θ ) sin − 1 ( sin θ ) = sin θ ,if θ ε [ − π 2 , π 2 ] y = θ y = cot − 1 x Differentiating with respect to x , We get
d y d x = d d x ( cot − 1 x ) ∴ d d x ( cot − 1 x ) ⇒ − 1 1 + x 2 d y d x = − 1 1 + x 2
Differentiation Exercise 10.3 Question 23
Answer:
d y d x = 2 n x n − 1 1 + x 2 n Hint:
d dx ( constsant ) = 0 d d x ( x n ) = n x n − 1 Given:
cos − 1 { 1 − x 2 n 1 + x 2 n } 0 < x < ∞ Solution:
y = c o s − 1 { 1 − x 2 n 1 + x 2 n } Let,
x n = tan θ θ = tan − 1 ( x n ) y = cos − 1 { 1 − tan 2 θ 1 + tan 2 θ } y = cos − 1 { cos 2 θ } Using,
1 − tan 2 θ 1 + tan 2 θ = cos 2 θ
Considering the limits,
0 < x < ∞ 0 < x n < ∞ 0 < θ < π 2
Now, { cos − 1 ( cos θ ) = θ , i f θ ε [ 0 , π ] }
y = cos − 1 ( cos 2 θ )
y = 2 θ S i n c e x n = tan θ d y d x = d d x ( 2 tan − 1 ( x n ) ) d d x ( tan − 1 x ) = 1 1 + x 2 d y d x = 2 × 1 1 + ( x n ) 2 × n x n − 1 d y d x = 2 n x n − 1 1 + x 2 n
Differentiation Exercise 10.3 Question 24
Answer:
d y d x = 0 Hint:
d dx ( constsant ) = 0 d d x ( x n ) = n x n − 1 Given:
sin − 1 ( 1 − x 2 1 + x 2 ) + sec − 1 ( 1 + x 2 1 − x 2 ) Solution:
Let,
y = sin − 1 ( 1 − x 2 1 + x 2 ) + sec − 1 ( 1 + x 2 1 − x 2 ) Using
sec − 1 x = cos − 1 ( 1 x ) y = sin − 1 ( 1 − x 2 1 + x 2 ) + cos − 1 ( 1 − x 2 1 + x 2 ) Using
sin − 1 x + cos − 1 x = π 2 y = π 2 Differentiating with respect to x we get
d y d x = d d x ( π 2 )
d y d x = 0 { d d x ( c o n s t a n t ) = 0 }
Differentiation Exercise 10.3 Question 26
Answer:
1 2 x ( 1 + x ) Hint:
d d x ( c o n s t a n t ) = 0 ; d d x ( x n ) = n x n − 1 Given:
tan − 1 ( x + a 1 − x a ) Solution:
Let,
y = tan − 1 ( x + a 1 − x a ) Since,
tan − 1 x + tan − 1 y = tan − 1 ( x + y 1 − xy ) y = tan − 1 x + tan − 1 a Differentiating it with respect to x d y d x = d d x ( tan − 1 x ) + d d x ( tan − 1 a ) Since , d dx ( tan − 1 x ) = 1 1 + x 2 ; d dx ( x n ) = n x n − 1 dy dx = 1 1 + ( x ) 2 d dx ( x ) + 0 dy dx = 1 1 + ( x 1 / 2 ) 2 × 1 2 ( x ) 1 2 − 1 dy dx = 1 1 + x × 1 2 ( x ) − 1 2 dy dx = 1 1 + x × 1 2 x dy dx = 1 2 x ( 1 + x ) Differentiation Exercise 10.3 Question 27
Answer:
d y d x = 1 Hint:
d d x ( c o n s t a n t ) = 0 d d x ( x n ) = n x n − 1 Given:
tan − 1 [ a + btan x b − atan x ] Solution:
Let,
y = tan − 1 [ a + b tan x b − atan x ] y = tan − 1 [ a + b tan x b b − a t a n x b ] y = tan − 1 [ a b + tan x 1 − a b tan x ] y = tan − 1 [ tan ( tan − 1 a b ) + tan x 1 − tan ( tan − 1 ⋅ g b ) tan x ] y = tan − 1 [ tan ( tan − 1 a b + x ) ] ∴ tan ( A + B ) = tan A + tan B → 1 − tan A tan B y = tan − 1 ( a b ) + x Differentiating it with respect to x , We get d y d x = d d x ( tan − 1 ( a b ) ) + d d x ( x ) d d x ( constant ) = 0 d d x ( tan − 1 x ) = 1 1 + x 2 d y d x = 1 1 + ( a b ) 2 × 0 + 1 d y d x = 0 + 1 d y d x = 1 Differentiation Exercise 10.3 Question 28
Answer:
d y d x = 1 1 + x 2 Hint:
d d x ( constant ) = 0 ; d d x ( x n ) = n x n − 1 Given:
tan − 1 ( a + b x b − a x ) Solution:
Let,
y = tan − 1 ( a + b x b − a x ) y = tan − 1 [ a + b x b b − a x b ] Since,
tan − 1 x + tan − 1 y = tan − 1 ( x + y 1 − x y ) Differentiating it with respect to x , We get
d y d x = d d x ( tan − 1 ( a b ) + tan − 1 x Since, d d x ( tan − 1 x ) = 1 1 + x 2 d y d x = 1 1 + ( a b ) 2 × 0 + 1 1 + x 2 d y d x = 0 + 1 1 + x 2 d y d x = 1 1 + x 2
Differentiation Exercise 10.3 Question 29
Answer:
d y d x = a a 2 + x 2 Hint:
d d x ( constan t ) = 0 ; d d x ( x n ) = n x n − 1 Given:
tan − 1 ( x − a x + a ) Solution:
Let,
y = tan − 1 ( x − a x + a ) y = tan − 1 ( x − a x x + a x ) y = tan − 1 ( x x − a x x x + a x ) y = tan − 1 ( 1 − a x 1 + 1 × a x ) y = tan − 1 ( 1 ) + tan − 1 ( a x ) Using,
tan − 1 x + tan − 1 y = tan − 1 ( x + y 1 − x y )
Differentiating its with respect to x using chain tule.
d y d x = 0 − 1 1 + ( a x ) 2 d d x ( a x ) d y d x = − 1 1 + a 2 x 2 ( − a x 2 ) d y d x = − 1 x 2 + a 2 x 2 × ( − a x 2 ) d y d x = − x 2 x 2 + a 2 × ( − a x 2 ) d d x ( x n ) = n x n − 1
d d x = a x 2 + a 2 d d x ( c o n s t a n t ) = 0
Differentiation Exercise 10.3 Question 30
Answer:
d y d x = 3 1 + 9 x 2 − 2 1 + 4 x 2 Hint:
d dx ( constan t ) = 0 ; d d x ( x n ) = n x n − 1 Given:
tan − 1 ( x 1 + 6 x 2 ) Solution:
Let,
y = tan − 1 ( x 1 + 6 x 2 ) y = tan − 1 ( 3 x − 2 x 1 + ( 3 x ) ( 2 x ) ) Since,
3 x − 2 x = x y = tan − 1 3 x − tan − 1 2 x [ Since + tan x − tan − 1 y = tan − 1 ( x − y 1 + x y ) ] Differentiating it with respect tox using chain rule, d y d x = 1 1 + ( 3 x 2 ) d d x ( 3 x ) − 1 1 + ( 2 x ) 2 d d x ( 2 x ) { Since ⇒ d d x tan − 1 ( x ) = 1 1 + x 2 } d y d x = 1 1 + 9 x 2 ( 3 ) − 1 1 + 4 x 2 ( 2 ) d y d x = 3 1 + 9 x 2 − 2 1 + 4 x 2 Differentiation Exercise 10.3 Question 31
Answer:
d y d x = 3 1 + 9 x 2 + 2 1 + 4 x 2 Hint:
d d x ( c o n s t a n t ) = 0 ; d d x ( x n ) = n x n − 1 Given:
tan − 1 ( 5 x 1 − 6 x 2 ) Solution:
Let,
y = tan − 1 ( 5 x 1 − 6 x 2 ) y = tan − 1 ( 3 x + 2 x 1 − ( 3 x ) ( 2 x ) ) By using 5 x = 3 x + 2 x y = tan − 1 ( 3 x ) + tan − 1 ( 2 x ) Since, tan − 1 x + tan − 1 y = tan − 1 ( x + y 1 − x y ) Differentiating it with respect to x using chain rule, d y d x = d d x ( tan − 1 ( 3 x ) ) + d d x ( tan − 1 ( 2 x ) ) Using d d x ( tan − 1 x ) = 1 1 + x 2 d y d x = 1 1 + ( 3 x ) 2 d d x ( 3 x ) + 1 1 + ( 2 x ) 2 d d x ( 2 x ) d y d x = 1 1 + 9 x 2 ( 3 ) + 1 1 + 4 x 2 ( 2 ) d y d x = 3 1 + 9 x 2 + 2 1 + 4 x 2 Differentiation Exercise 10.3 Question 32
Answer:
d y d x = 1 Hint:
d dx ( constan t ) = 0 ; d d x ( x n ) = n x n − 1 Given:
tan − 1 [ cos x + sin x cos x − sin x ] Solution:
y = tan − 1 [ cos x + sin x cos x − sin x ] y = tan − 1 [ cos x + sin x cos x cos x − sin x cos x ] y = tan − 1 [ 1 + tan x 1 − tan x ] since, sin x cos x = tian x y = tan − 1 [ tan π 4 + tan x 1 − tan π 4 tan x ] y = tan − 1 [ tan ( π 4 + x ) ] Since tan ( A + B ) = tan A + tan B 1 − tan tan B y = π 4 + x Differentiating it with respect to x, d y d x = 0 + 1 { d dx ( constant ) = 0 d dx ( x ) = 1 } d y d x = 1 Differentiation Exercise 10.3 Question 33
Answer:
d y d x = 1 3 x 2 ( 1 + x 2 3 ) Hint:
d d x ( constant ) = 0 ; d d x ( x n ) = n x n − 1 Given:
tan − 1 [ x 1 / 3 + a 1 / 3 1 − ( ax ) 1 / 3 ] Solution:
Let,
y = tan − 1 [ x 13 + a 1 / 3 1 − ( a x ) 1 / 3 ] y = tan − 1 ( x 1 / 3 ) + tan − 1 ( a 1 / 3 ) Since, tan − 1 x + tan − 1 y = tan − 1 ( x + y 1 − x y ) Differentiating it with respect to x using chain rule, d y d x = 1 1 + ( x 1 / 3 ) 2 d d x ( x 1 / 3 ) + 1 1 + ( a 1 / 3 ) 2 d d x ( a 1 / 3 ) d y d x = 1 1 + x 2 / 3 × ( 1 3 x 1 3 − 1 ) + 1 1 + ( a 1 / 3 ) 2 × 0 d y d x = 1 1 + x 2 / 3 × 1 3 x − 2 / 3 d y d x = 1 3 x 2 / 3 ( 1 + x 2 / 3 ) Using d d x ( tan − 1 x ) = 1 1 + x 2 d d x ( x n ) = n x n − 1 d d x ( constant ) = 0 Differentiation Exercise 10.3 Question 34
Answer:
d y d x = 2 x + 1 l o g 2 1 + 4 x Hint:
d d x ( c o n s t a n t ) = 0 ∂ ∂ x ( x n ) = n x n − 1 Given:
sin − 1 ( 2 x + 1 1 + 4 x ) Solution:
Let
y = sin − 1 ( 2 x + 1 1 + 4 x ) To find the domain we need to find all x − 1 ≤ 2 x + 1 1 + 4 x ≤ 1 Since the quantity in the middle is always positive, we need to find
all such that
2 x + 1 1 + 4 x ≤ 1 ie, all
x such that
2 x + 1 = 1 + 4 x 2 ≤ 1 2 x + 2 x , which is true for a
x Hence the function is defined at all real numbers
Putting
2 x = tan θ y = sin − 1 ( 2 x + 1 1 + 4 x ) y = sin − 1 ( 2 x ⋅ 2 1 + ( 2 x ) 2 ) y = sin − 1 ( 2 tan θ 1 + tan 2 θ ) Using sec 2 θ = 1 + tan 2 θ y = sin − 1 ( 2 tan θ sec 2 θ ) y = sin − 1 ( 2 sin θ cos θ × cos 2 θ ) \; S i n c e , tan θ = sin θ cos θ ′ sec θ , 1 sec = cos θ y = sin − 1 ( 2 sin θ cos θ ) U sin sin = sin 2 θ = 2 sin θ cos θ y = sin − 1 ( sin θ ) { sin − 1 ( sin θ ) = θ if θ ∈ [ − 1 2 , 0 2 ] } y = 2 θ y = 2 tan − 1 ( 2 x ) { since 2 x = tan θ θ = tan − 1 ( 2 x ) } d y d x ⇒ d d x ( 2 tan − 1 ( 2 x ) ) d y d x = 2 × 1 1 + ( 2 x ) 2 ∂ d x ( 2 x ) d y d x = 2 × 1 1 + 4 x ⋅ ( 2 x ) log 2 d y d x ⇒ 2 x + 1 log 2 1 + 4 x Since d d x ( tan − 1 x ) = 1 1 + x 2 Differentiation Exercise 10.3 Question 35
Answer: Hence Prove
d y d x = 4 1 + x 2 Hint:
d d x ( constan t ) = 0 ; d d x ( x n ) = nx n − 1 Given:
y = sin − 1 ( 2 x 1 + x 2 ) + sec − 1 ( 1 + x 2 1 − x 2 ) 0 < x < 1 Solution:
Prove :
d y d x = 4 1 + x 2 y = sin − 1 ( 2 x 1 + x 2 ) + cos − 1 ( 1 − x 2 1 + x 2 ) Since sec − 1 x = cos − 1 ( 1 x ) y = sin − 1 ( 2 x 1 + x 2 ) + cos − 1 ( 1 − x 2 1 + x 2 ) Let,
x = tan θ θ = tan − 1 x y = sin − 1 ( 2 tan θ 1 + tan 2 θ ) + cos − 1 ( 1 − tan 2 θ 1 + tan 2 θ ) y = sin − 1 ( 2 tan θ sec 2 θ ) + cos − 1 ( cos 2 θ ) Using
sec 2 θ = 1 + tan 2 θ 1 sec θ = cos θ 1 − tan 2 θ 1 + tan 2 θ = cos 2 θ y = sin − 1 ( 2 sin θ cos θ × cos 2 θ ) + cos − 1 ( Cos 2 θ ) y = sin − 1 ( 2 sin θ cos θ ) + cos − 1 ( cos 20 ) y = sin − 1 ( sin 2 θ ) + cos − 1 ( cos 2 θ ) − ( i ) u sing 2 sin θ cos θ = sin 2 θ Considering Limits, 0 < x < 1 0 < tan θ < 1 0 < θ < π 4 0 < ( 2 θ ) < π 2 so from e i ( i ) y = 2 θ + 2 θ y = 4 θ cos − 1 ( cos θ ) ⇒ if θ ∈ [ 0 , π ] y = 4 tan − 1 x s i n c e , x = tan θ , θ = tan − 1 x Differentiating it with respect to x d y d x = 4 d d x ( tan − 1 x ) Using d d x ( tan − 1 x ) = 1 1 + x 2 d y d x = 4 ( − 1 1 + x 2 ) d y d x = 4 1 + x 2 Differentiation Exercise 10.3 Question 36
Answer: Hence Prove ,
d y d x = 2 1 + x 2 Hint:
d d x ( constan t ) = 0 ; d d x ( x n ) = n x n − 1 Given:
y = sin − 1 ( x 1 + x 2 ) + cos − 1 ( 1 1 + x 2 ) 0 < x < ∞ Solution:
Let,
x = tan θ , θ = tan − 1 x y = sin − 1 ( tan θ 1 + tan 2 θ ) + cos − 1 ( 1 1 + tan 2 θ ) y = sin − 1 ( tan θ sec 2 θ ) + cos − 1 ( 1 sec 2 θ ) Using sec 2 θ = 1 + tan 2 θ sin θ cos θ = tan θ 1 sec θ = θ cos θ y = sin − 1 ( tan θ sec θ ) + cos − 1 ( 1 sec θ ) y = sin − 1 ( sin θ cos θ × cos θ ) + cos − 1 ( cos θ ) y = sin − 1 ( sin θ ) + cos − 1 ( cos θ ) − ( i ) Considering limit 0 < x < ∞ 0 < tan θ < ∞ 0 < θ < π 2 So From eq (i)
y = θ + θ Since, sin − 1 ( sin θ ) = θ if θ ε [ − π 2 , π 2 ] y = 2 θ cos − 1 ( cos θ ) = θ if θ ε [ 0 , π ] y = 2 tan − 1 x [ s i n c e x = tan θ ] Differentiating it with respect to x
d d x ( tan − 1 x ) = 1 1 + x 2 d y d x = 2 × 1 1 + x 2 d y d x = 2 1 + x 2
Differentiation Exercise 10.3 Question 37(i)
Answer:
d y d x = − 1 Hint:
d dx ( constan t ) = 0 d d x ( x n ) = n x n − 1 Given:
cos − 1 ( sin x ) Solution:
Let,
y = cos − 1 ( sin x ) We observe that this function is defined for all real numbers y = cos − 1 ( sin x ) y = cos − 1 [ cos ( π 2 − x ) ] y = π 2 − x { since cos − 1 ( cos θ ) = θ if θ ∈ [ θ , π ] } Differentiating it with respect to x, d y d x = 0 − 1 d y d x = − 1 { S i n c e , d ( c o n s t a n t ) d x = 0 } Differentiation Exercise 10.3 Question 37 (ii)
Answer:
d y d x = 1 1 + x 2 Hint:
d d x ( constan t ) = 0 d d x ( x n ) = n x n − 1 Given:
cos − 1 ( 1 − x 1 + x ) Solution:
y = cos − 1 ( 1 − x 1 + x ) Let,
x = tan θ θ = tan − 1 x y = cot − 1 ( 1 − tan θ 1 + tan θ ) y = cot − 1 ( tan n 4 − tan θ 1 + tan n 4 tan θ ) y = cot − 1 [ tan ( n 4 − θ ) ] Using, tan ( A − B ) = tan A − tan B 1 + tan A tan B y = cot − 1 [ tan ( π 4 − θ ) ] y = cot − 1 [ cots ( π 2 − ( π − θ 4 − θ ) ] ∴ s i n c e cot ( π 2 − θ ) = tan θ y = cot − 1 [ cot ( π 2 − π 4 + θ ) ] y = cot − 1 [ cot ( π 4 + θ ) ] y = π 4 + θ { cot − 1 ( cot θ ) = θ , if θ < [ − π 2 , π 2 ] } y = π 4 + tan − 1 x { Since x = tan θ θ = tan − 1 x } Differentiating it with respect to x, d y d x = 0 + 1 1 + x 2 d d x ( tan − 1 x ) = 1 1 + x 2 ∂ d x ( constant ) = 0 d y d x = 1 1 + x 2 Differentiation Exercise 10.3 Question 38
Answer:
d y d x = 1 2 , Hence,
d y d x is independent of x
Hint:
d d x (constant) = 0 ; d d x ( x n ) = n x n − 1 Given:
y = cot − 1 { 1 + sin x + 1 − sin x 1 + sin x − 1 − sin x } Solution:
y = cot − 1 { 1 + sin x + 1 − sin x 1 + sin x − 1 − sin x } Then, 1 + sin x + 1 − sin x 1 + sin x − 1 − sin x 1 + sin x + 1 − sin x 1 + sin x − 1 − sin x × 1 + sin x + 1 − sin x 1 + sin x + 1 − sin x ( 1 + sin x + 1 + sin x ) 2 ( 1 + sin x ) − ( 1 − sin x ) Using ( a − b ) ( a + b ) = a 2 − b 2 ( a + b ) 2 = a 2 + b 2 + 2 a b ( 1 + sin x ) + ( 00 ( 1 − sin x ) + 2 ( 1 − sin x ) ( 1 + sin x ) ( 1 + sin x ) − ( 1 − sin x ) 1 + sin x + 1 − sin x + 2 ( 1 − sin x ) ( 1 + sin x ) 1 + sin x − 1 + sin x ⇒ 2 + 2 1 + sin x − sin x − sin 2 x 2 sin x ⇒ 2 + 2 1 − sin 2 x 2 sin x ⇒ 2 ( 1 + 1 − sin 2 x ) 2 sin x ⇒ 1 + 1 sin 2 x sin x Using
cos 2 x + sin 2 x = 1 ⇒ 1 + cos 2 x sin x ⇒ 1 + cos x sin x Using sin 2 θ = 2 sin θ cos θ 1 + Cos 2 θ = 2 cos 2 θ ⇒ 2 cos 2 x 2 2 sin x 2 cos x 2 ⇒ cot x 2 { cos x sin x = cot x } Therefore, equation (1) becomes
y = cot − 1 ( cot x 2 ) y = x 2 { cot − 1 ( cot θ ) = θ if θ ε [ − π 2 , π 2 ] }
Differentiating it with respect to x
d y d x = 1 2 d d x ( x )
d y d x = 1 2
Differentiation Exercise 10.3 Question 39
Answer:
d y d x = 4 1 + x 2 Hint:
d dx ( constiant ) = 0 d d x ( x n ) = nx n − 1 Given:
y = tan − 1 ( 2 x 1 − x 2 ) + sec − 1 ( 1 + x 2 1 − x 2 ) x = 0 Prove
d y d x = 1 1 + x 2 Solution:
Here,
$y=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\sec ^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right) $ Using
sec − 1 x = cos − 1 ( 1 x ) y = tan − 1 ( 2 x 1 − x 2 ) + cos − 1 ( 1 − x 2 1 + x 2 ) Put
x = tan θ y = tan − 1 ( 2 tan θ 1 − tan 2 θ ) + cos − 1 ( 1 − tan 2 θ 1 + tan 2 θ ) y = tan − 1 ( tan 2 θ ) + cos − 1 ( cos 2 θ ) − ( 1 ) Using,
tan 2 θ = 2 tan θ 1 − tan 2 θ
1 − tan 2 θ 1 + tan 2 θ = cos 2 θ Here, 0 < x < ∞ 0 < tan θ < ∞ 0 < θ < π 2 0 < 2 θ < π
So From eq(i)
y = 2 θ + 2 θ
{ Since, tan − 1 ( tan θ ) = θ if θ ∈ [ − π 2 , π 2 ] and cos − 1 ( cos θ ) = θ if θ [ 0 , π ] }
y = 4 θ
y = 4 tan − 1 x [ U s i n g x = tan θ , θ = tan − 1 x ]
Differentiating it with respect t0 x
d y d x = 4 1 + x 2 ⇒ 4 1 + x 2 Using d ( tan − 1 x ) d x = 1 1 + x 2
Differentiation Exercise 10.3 Question 40
Answer:
d y d x = 0 Hint:
d d x ( x n ) = n x n − 1 ; d dx ( constant ) = 0 Given:
y = sec − 1 ( x + 1 x − 1 ) + sin − 1 ( x − 1 x + 1 ) x > 0 Solution:
y = sec − 1 ( x + 1 x − 1 ) + sin − 1 ( x − 1 x + 1 ) x > 0 y = cos − 1 ( x − 1 x + 1 ) + sin − 1 ( x − 1 x + 1 ) Using
sec − 1 x = cos ( 1 x ) Since,
cos − 1 x + sin − 1 x = π 2 y = π 2 Differentiating with respect to x d y d x = 0 [ d d x ( c o n s t a n t ) = a ] ] Differentiation Exercise 10.3 Question 41
Answer:
d y d x = − x 1 − x 2 Hint:
d dx ( constant ) = 0 ; d dx ( x n ) = n x n − 1 Given:
y = sin [ 2 tan − 1 { x 1 + x } ] Solution:
Put
x = cos 2 θ y = sin [ 2 tan − 1 1 − cos 2 θ 1 + cos 2 θ ] y = sin [ 2 tan − 1 2 sin 2 θ 2 cos 2 θ ] Since 1 − cos 2 θ = 2 sin 2 θ 1 + cos 2 θ = 2 cos 2 θ 1 + cos 2 θ = 2 cos 2 θ y = sin [ 2 tan − 1 tan 2 θ ] Since sin θ cos θ = tan θ y = sin [ 2 tan − 1 ( tan θ ) ] y = sin [ 2 θ ] { tan − 1 ( tan θ ) = θ i f θ ε [ − π 2 , π 2 ] } y = sin ( 2 × 1 2 cos − 1 x ) [ S i n c e , x = cos 2 θ , θ = 1 2 cos − 1 x ] y = sin ( cos − 1 x ) y = sin ( sin − 1 1 − x 2 ) y = 1 − x 2 [ sin − 1 ( sin θ ) ⇒ 0 if θ ∈ [ − π 2 , π 2 ] ] Differentiating it with respect to x d y d x = d d x ( 1 − x 2 ) = d d x ( 1 − x 2 ) 1 / 2 As we know, d d x ( x n ) = n x n − 1 d y d x = 1 2 ( 1 − x 2 ) 1 2 d ∂ x ( 1 − x 2 ) d y = 1 d x = 1 2 1 − x 2 ( − 2 x ) d y d x = − x 1 − x 2 Differentiation Exercise 10.3 Question 42
Answer:
d y d x = 2 1 − 4 x 2 Hint:
d d x (constant) = 0 d d x ( x n ) = nx − 1 Given:
y = cos − 1 ( 2 x ) + 2 cos − 1 1 + 4 x 2 , 0 < x < 1 2 Solution:
Put
2 x = cos θ So,
y = cos − 1 ( cos θ ) + 2 cos − 1 1 − cos 2 θ Since cos 2 θ + sin 2 θ = 1 y = cos − 1 ( cos θ ) + 2 cos − 1 sin 2 θ y = cos − 1 ( cos θ ) + 2 cos − 1 ( sin θ ) y = cos ( cos θ ) + 2 cos − 1 ( cos ( π 2 − θ ) ) − ( i ) 0 < x = 1 0 < 2 x < 1 0 < cos θ = 1 0 < θ < π 2 And 0 > − θ > − π 2 n 2 > ( n 2 − θ ) > 0 So From eq (i)
y = θ + 2 ( π 2 − θ ) { S i n c e cos − 1 ( cos θ ) = 0 i f θ ε [ 0 , n ] } y ⇒ θ + π − 2 θ y ⇒ π − C y = π − cos − 1 ( 2 x ) Since, 2 x = cos θ Differentiating it with respect to x d y d x = 0 − [ − 1 1 − ( 2 x ) 2 ] d d x ( 2 x ) Since,
d d x ( c o n s t a n t ) = 0 ; d d x ( cos − 1 x ) = 1 1 − x 2 d y d x = 1 1 − 4 x 2 ( 2 ) d y d x = 2 1 − 4 x 2 Differentiation Exercise 10.3 Question 43
Answer: Hence Prove ,
1 + a 2 = b Hint:
d d x ( x n ) = n x n − 1 d ( constant) d x = 0 Given:
d dx [ tan − 1 ( a + bx ) ] = 1 at x = 0 Solution:
So, using chain rule, we know d d x ( tan − 1 x ) = 1 1 + x 2 [ { 1 1 + ( a + b x ) 2 } d d x ( a + b x ) ] x = 0 = 1 [ 1 1 + ( a + b x ) 2 × ( b ) ] x = 0 = 1 [ b [ 1 + ( a + b x ) 2 ] x = 0 = 1 ∴ ( a + b ) 2 = a 2 + b 2 + 2 a b [ b 1 + ( a 2 + b 2 x 2 + 2 a b x ) ] x = 0 = 1 b 1 + ( a 2 + 0 + 0 ) = 1 b 1 + a 2 = 1 b = 1 + a 2 Hence Proved, Differentiation Exercise 10.3 Question 44
Answer:
d y d x = − 6 1 − 4 x 2 Hint:
d d x ( x n ) = n x n − 1 d d x ( Constant ) = 0 Given:
Solution:
y = cos − 1 ( 2 x ) + 2 cos − 1 1 − 4 x 2 , − 1 2 < x < 0 Put
2 x = cos θ So
y = cos − 1 ( coc θ ) + 2 cos − 1 1 − cos 2 θ Using,
sin 2 θ + cos 2 θ = 1 y = cos − 1 ( cos θ ) + 2 cos sin 2 θ y = cos − 1 ( cos θ ) + 2 cos − ( sin θ ) y = cos − 1 ( cos θ ) + 2 cos ( cos − 1 ( π 2 − θ ) ) − ( i ) Considering limit, − 1 2 < x < 0 − 1 < 2 x < 0 − 1 − cos θ < 0 π 2 < θ < π And, − π 2 > − θ > − π ( π 2 − π 2 ) > ( π 2 − θ ) > ( π 2 − π ) 0 > ( π 2 − θ ) > − π 2 So, from equation (i) y = θ + 2 [ − ( π 2 − θ ) ] { Since , cos − 1 ( cos θ ) = 0 if θ ε [ 0 , π ] , cos − 1 cos ( θ ) = − θ , if θ ε [ − π ⋅ 0 ] } y = θ − 2 × π 2 + 2 θ y = − π + 3 θ y = − π + 3 cos − 1 ( 2 x ) Differentiating its with respect to x
d y d x = 0 + 3 [ − 1 1 − ( 2 x ) 2 ] d d x ( 2 x )
As we Know,
d dx ( constant ) = 0 d dx ( cos − 1 x ) = − 1 1 − x 2 dy dx = − 3 1 − 4 x 2 − ( 2 ) dy dx = − 6 1 − 4 x 2
Differentiation Exercise 10.3 Question 45
Answer:
d y d x = 1 2 1 − x 3 Hint:
d d x ( constant ) = 0 d d x ( x n ) = n x n − 1 Given:
y = tan − 1 { 1 + x − 1 − x 1 + x + 1 − x } Solution:
Put,
x = cos 2 θ So,
y = tan − 1 [ 1 + cos 2 θ − 1 − cos 2 θ 1 + cos 2 θ + 1 − cos 2 θ ] = tan − 1 ( 2 cos 2 θ − 2 sin 2 θ 2 cos 2 θ + 2 sin 2 θ ) Using
1 + cos 2 θ = 2 cos 2 θ y = tan − 1 ( 2 cos θ − 2 sin θ 2 cos θ + 2 sin θ ) y = tan − 1 ( 2 ( cos θ − sin θ ) 2 ( cos θ + sin θ ) ) Dividing numerator and denominator by cos θ
y = tan − 1 ( cos θ cos θ − sin θ cos θ cos θ cos θ + sin θ cos θ ) y = tan − 1 ( cos θ − sin θ cos θ cos θ + sin θ cos θ )
Using
tan θ = sin θ cos θ y = tan − 1 ( 1 − tan θ 1 + tan θ ) y = tan − 1 ( tan n 4 − tan θ 1 + tan π 4 tan θ ) y = tan − 1 ( tan ( n 4 − θ ) ) Using tan ( A − B ) = tan A − tan B 1 + tan A tan B
y = π − 0 4 − θ [ U s i n g tan − 1 ( tan θ ) = θ i f θ ε [ − π 2 , π 2 ] ]
y = π 4 − 1 2 cos − 1 x [ Using x = cos 2 θ ] dy d x = 0 − 1 2 ( − 1 1 − x 2 ) { d d x ( cos − 1 x ) = − 1 1 − x 2 } d y d x = 1 2 1 − x 2 d y d x = − 1 1 − x 2
Differentiation Exercise 10.3 Question 47
Answer:
d y d x = − 1 1 − x 2 Hint:
d dx ( constsant ) = 0 d dx ( x n ) = n x n − 1 Given:
y = cos − 1 { 2 x − 3 1 − x 2 13 ) Solution:
Let,
x = cos θ y = cos − 1 { 2 cos θ − 3 1 − cos 2 θ 4 } Using cos 2 θ + sin 2 θ = 1 y = cos − 1 { 2 13 cos θ − 3 13 sin 2 θ ] y = cos − 1 { 2 13 cos θ − 3 13 sin θ ] cos ϕ = 2 13 Let,
sin ϕ = 1 − cos 2 ϕ = 1 − ( 2 13 ) 2 + 1 − 4 13 = 13 − 4 13 = 9 13 sin ϕ = 3 13 So,
y = cos − 1 { cos ϕ cos θ − sin ϕ sin θ ] = cos − 1 [ cos ( θ + ϕ ) ] Using, cos ( A + B ) = cos A cos B − sin A sin B y = ϕ + θ [ cos − 1 ( cos θ ) = θ if 0 ε [ 0 , π ] ] y = cos − 1 ( 2 13 ) + cos − 1 x since, x = cos θ cos ϕ = 2 13 Differentiating its with Respect to x d y d x = − 1 1 − ( 2 13 ) 2 d d x ( 2 13 ) + ( − 1 1 − x 2 ) d y d x = − 1 1 − ( 2 13 ) 2 × 0 + ( − 1 1 − x 2 ) d d x ( cos − 1 x ) = − 1 1 − x 2 d y d x = 0 − 1 1 − x 2 Differentiation Exercise 10.3 Question 48
Answer:
d y d x = 6 1 − 9 x 2 Hint:
d dx (constant ) = 0 d dx ( x n ) = n x n − 1 Given:
y = sin − 1 ( 6 x 1 − 9 x 2 ) 1 3 2 < x < 1 3 2 Solution:
Let
x = 1 3 sin θ y = sin − 1 ( 6 × 1 3 sin θ 1 − 9 ( 1 3 ) 2 sin 2 θ ) y = sin − 1 ( 2 sin θ 1 − 9 × 1 9 sin 2 θ ) y = sin − 1 ( 2 sin θ 1 − sin 2 θ ) U sin g sin 2 θ + cos 2 θ = 1 y = sin − 1 ( 2 sin θ cos 2 θ ) y = sin − 1 ( 2 sin θ cos θ ) y = sin − 1 ( sin 2 θ ) − ( i ) U sin g sin 2 θ = 2 sin θ cos θ Considering limits here 1 3 2 < x < 1 3 2 1 3 × 1 3 2 < 1 3 sin θ < 1 3 2 × 1 3 1 9 2 < 1 3 sin θ < 1 9 2 From Equation (i)
y = 2 θ { sin − 1 ( sin θ ) = θ if θ ε [ − π 2 , π 2 ] }
y = 2 sin − 1 ( 3 x )
Differentiating it with respect tox
d y d x = d d x ( 2 sin − 1 3 x ) As we thow, d d x ( sin − 1 x ) = 1 1 − x 2 d y d x = 2 × 1 1 − ( 3 x ) 2 d d x ( 3 x ) d y d x = 2 1 − 9 x 2 × 3 d y d x = 6 1 − 9 x 2
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It is a well-known fact that NCERT is a book that every teacher follows. Keeping this fact in mind, the students are provided with solutions that abide by NCERT and CBSE guidelines. Rd Sharma Class 12th Exercise 10.3 solutions help students achieve good marks in-class exams also.
Alternative ways to solve a question
RD Sharma Class 12 Solutions Chapter 10 ex 10.3 because they are crafted by experts who have many ways to solve a single question. Every expert has a different way to solve a problem, here also a student will find many forms and can choose what suits them best.
Great performance
These solutions help a student cross their benchmark scores in the exams. The Class 12th RD Sharma Chapter 10.3 Exercise 10.3 Solutions includes all the concepts that are present in the textbook.
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RD Sharma Chapter-wise Solutions
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