Class 12 Maths NCERT Solutions for Chapter 5 Continuity and Differentiability Exercise 5.3
![Class 12 Maths NCERT Solutions for Chapter 5 Continuity and Differentiability Exercise 5.3 Class 12 Maths NCERT Solutions for Chapter 5 Continuity and Differentiability Exercise 5.3](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjXGxpGiXSjqC0_GBwV2gFeJTzCcL1NDoqkqFkmlKXG2PMg0JvNw_lvjikoF9tGAnGP3HqmU0Z-mcbupZMDc3l9Z_EE3S19hdpN3H_yN1VdDHx5BOkKWQPr4J7Q0Jun430K-Yn3IqhOUTH0z0HtvSVNmkixAtSkU2Mzl4PhMPkoiImV553VLl-C0pOf/w650-h307-rw/cbse-class-12-maths-ncert-solutions-chapter-5-exercise-5-3.jpg)
Continuity and Differentiability Exercise 5.3 Solutions
1. Find dy/dx .
2x + 3y = sin x
Solution
The given relationship is 2x + 3y = sin x
Differentiating this relationship with respect to x, we obtain
2. Find dy/dx
2x + 3y = sin y
Solution
The given relationship is 2x + 3y = sin y
Differentiating this relationship with respect to x, we obtain
3. Find dy/dx
ax + by2 = cos y
Solution
The given relationship is ax + by2 = cos y
Differentiating this relationship with respect to x, we obtain
4. Find dy/dx
xy + y2 = tan x + y
Solution
The given relationship is xy + y2 = tan x + y
Differentiating this relationship with respect to x, we obtain
5. Find dy/dx
x2 + xy + y2 = 100
Solution
The given relationship is x2 + xy + y2 = 100
Differentiating this relationship with respect to x, we obtain
x3 + x2 y + xy2 + y3 = 81
Solution
The given relationship is x3 + x2 y + xy2 + y3 = 81
Differentiating this relationship with respect to x, we obtain
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh0E1WfE2LWEGBDmerKYYBO1kvsLoiHsR81gujqi3zuIYmyKtN7638A0jO7mfd8Gl0VMkhvbYLOpUGVOrPlkuezDbUrLqp9gsUwX7eRE5b64zQ8EbbPP17H10S1e5LV6qbSPap6yosoVHTd2IR5DiDARamyvJ6Ip1R_tPSi-RVKdknXcs8rLJQ6hso3/w480-h317-rw/Class%2012%20NCERT%20Solution%20Chapter%205%20Continuity%20and%20Differentiability%20Exercise%20-%205.3%20img%206.JPG)
Differentiating this relationship with respect to x, we obtain
7. Find dy/dx
sin2 y + cos xy = π
sin2 y + cos xy = π
Solution
The given relationship is sin2 y + cos xy = π
Differentiating this relationship with respect to x, we obtain
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjqJmeWxdzUZJZ9SbGuyoz1nqJzVHLnjONHH2ywgg1pG3WlW5lZEXfiAJoAdvwy0gEZ3xYifRjiw0oG-fRPjOvrn8r39Tjl4YUJtYQJGpOoARv36but0PoivuVtYNeC1LSMmvlSy-udR-BYAlupFAQ11FBo-eE-vjgGSdgPeJSKKX585cs4xqiDlWQL/w384-h265-rw/Class%2012%20NCERT%20Solution%20Chapter%205%20Continuity%20and%20Differentiability%20Exercise%20-%205.3%20img%207.JPG)
Differentiating this relationship with respect to x, we obtain
From (1), (2) and (3), we obtain
2 sin y cos y dy/dx - y sin xy - (x sin xy) dy/dx = 0
⇒ (2 sin y cos y - x sin xy) dy/dx = y sin xy
⇒ (sin 2y - x sin xy) dy/dx = y sin xy
∴ dy/dx = y sin xy/(sin 2y - x sin xy)
2 sin y cos y dy/dx - y sin xy - (x sin xy) dy/dx = 0
⇒ (2 sin y cos y - x sin xy) dy/dx = y sin xy
⇒ (sin 2y - x sin xy) dy/dx = y sin xy
∴ dy/dx = y sin xy/(sin 2y - x sin xy)
8. Find dy/dx
sin2 x + cos2 y = 1
sin2 x + cos2 y = 1
Solution
The given relationship is sin2 x + cos2 y = 1
Differentiating this relationship with respect to x, we obtain
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhekWQ8EF5W-6INSYxN1QDLSX1B_G-vlpx-5_R5Vy_Yu-jEgIbX947LWKbgSa6g4caF8zA8bdkiWKeeIivvsf-gKRfNpJbBrJE21lb_73vYyruL3Cp_ahOvKzdJ0MVFn3uifHNShZF59ZBDisPKD0bG5uB9YLvTi9l2SvdQ7C6OJIcTS8i3lsZVzP40/w301-h135-rw/Class%2012%20NCERT%20Solution%20Chapter%205%20Continuity%20and%20Differentiability%20Exercise%20-%205.3%20img%208.JPG)
Differentiating this relationship with respect to x, we obtain
⇒ 2 sin x cos x + 2cos y(-sin y). dy/dx = 0
⇒ sin 2x - sin 2y (dy/dx) = 0
∴ dy/dx = sin 2x/sin 2y
⇒ sin 2x - sin 2y (dy/dx) = 0
∴ dy/dx = sin 2x/sin 2y
9. Find dy/dx
y = sin-1 (2x/(1 + x2 )
y = sin-1 (2x/(1 + x2 )
Solution
The given relationship is y = sin-1 (2x/(1 + x2 )
y = sin-1 (2x/(1 + x2 )
⇒ sin y = 2x/(1 + x2 )
Differentiating this relationship with respect to x, we obtain
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiZl-4vlzUjoxTxIaUQX0of0PFLOYNL8JW41yG55Mr3J1NIHKfPb_VRJ59XQnaf9d-poM20uoZzYwdsWKVE4pqBxh2bLMwaI7d_x2IiwzCS5qxsLx3SVTitXXKZ0J8QdmVJON-HCVv7l2-4rLVWGcMGxWH8smADItIXO3l0Iee_Ks7BkaPvamtKMlK7/w338-h94-rw/Class%2012%20NCERT%20Solution%20Chapter%205%20Continuity%20and%20Differentiability%20Exercise%20-%205.3%20img%209.JPG)
The function, 2x/ (1 + x2 ) , is of the form of u/v.
Therefore, by quotient rule, we obtain
y = sin-1 (2x/(1 + x2 )
⇒ sin y = 2x/(1 + x2 )
Differentiating this relationship with respect to x, we obtain
The function, 2x/ (1 + x2 ) , is of the form of u/v.
Therefore, by quotient rule, we obtain
10. Find dy/dx
y = tan-1 (3x - x3 )/(1 - 3x2 ), -1/√3 < x < 1/√3
y = tan-1 (3x - x3 )/(1 - 3x2 ), -1/√3 < x < 1/√3
Solution
The given relationship is y = tan-1 (3x - x3 )/(1 - 3x2 ),
Comparing equations (1) and (2), we obtain
x = tan y/3
Differentiating this relationship with respect to x, we obtain
11. Find dy/dx
y = cos-1 (1 - x2 )/(1 + x2 ), 0 < x < 1
y = cos-1 (1 - x2 )/(1 + x2 ), 0 < x < 1
Solution
The given relationship is,
On comparing L.H.S. and R.H.S of the above relationship, we obtain
tan y/2 = x
Differentiating this relationship with respect to x, we obtain
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgvcqmpjMd3etTCRTWXWdLqkKueNE7rA20z9brEPCElq-1H6-zBK8n1TAOildAW_6uPGTXP5UEVRkvjUxDBYQ9QHcrfOIg0SCPbRLVmCR8BKZCDnqmIC3ItwfyK6teeY9gHSDKjTozQrQY2esVtKm2itZH7dB0qd_narprU7DnoXftcdyWosFT_ILAj/w168-h277-rw/Class%2012%20NCERT%20Solution%20Chapter%205%20Continuity%20and%20Differentiability%20Exercise%20-%205.3%20img%2014.JPG)
tan y/2 = x
Differentiating this relationship with respect to x, we obtain
12. Find dy/dx
y = sin-1 (1 - x2 )/(1 + x2 ), 0 < x < 1
Solution
The given relationship is y = sin-1 (1 - x2 )/(1 + x2 )
y = sin-1 (1 - x2 )/(1 + x2 )
⇒ sin y = (1 - x2 )/(1 + x2 )
Differentiating this relationship with respect to x, we obtain
d/dx (sin y) = d/dx [(1 - x2 )/(1 + x2 ) ] ...(1)
Using chain rule, we obtain
d/dx (sin y) = cos y . dy/dx
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiqoQi3UDmUIxqQz_amY4K61JTz7n7KxCknUJErCSuwzfRSl8xBkzXM_UL27A-tdT_lLDQnY6mC7KWB5ewBls-Xf-X_1eEPAD5s6hv-ElQjRQ7hZmV-7tv8i-c6RXO9JeUJRQZnCp-91UOMK9AeqZv98lxFIwKQA6Xz_a-HCwW9tCFuUKE5458AE9Ja/w543-h1374-rw/Class%2012%20NCERT%20Solution%20Chapter%205%20Continuity%20and%20Differentiability%20Exercise%20-%205.3%20img%2015.JPG)
⇒ sin y = (1 - x2 )/(1 + x2 )
Differentiating this relationship with respect to x, we obtain
d/dx (sin y) = d/dx [(1 - x2 )/(1 + x2 ) ] ...(1)
Using chain rule, we obtain
d/dx (sin y) = cos y . dy/dx
13. Find dx/dy
Solutiony = cos-1 (2x)/(1 + x2 ), 0 < x < 1
The given relationship is y = cos-1 (2x)/(1 + x2 )
y = cos-1 (2x)/(1 + x2 )
⇒ cos y = 2x/(1 + x2 )
Differentiating this relationship with respect to x, we obtain
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjxd1YjN_R7bKWQ80D8R3T9lu0KuegFmFXmsvXya181WH9lX9eZ0xzyVOYdr7Y4llzxIunFsIwvJjGxuZ-4UsmMh6j9gzkL5AvvbBQgIPapcclUnG96S09mBdTpwmEvy2tflAjg5YdYguQ0skwMtDMnO7P5hQSW-J2twW6P2_mKl7MAKOeRYlQP9BK6/w350-h580-rw/Class%2012%20NCERT%20Solution%20Chapter%205%20Continuity%20and%20Differentiability%20Exercise%20-%205.3%20img%2016.JPG)
y = cos-1 (2x)/(1 + x2 )
⇒ cos y = 2x/(1 + x2 )
Differentiating this relationship with respect to x, we obtain
14. Find dy/dx
y = sin-1 [2x√(1 - x2 )], -1/√2 < x < 1/√2
y = sin-1 [2x√(1 - x2 )], -1/√2 < x < 1/√2
Solution
The given relationship is y = sin-1 [2x√(1 - x2 )]
y = sin-1 [2x√(1 - x2 )]
⇒ sin y = 2x√(1 - x2 )
Differentiating this relationship with respect to x, we obtain
y = sin-1 [2x√(1 - x2 )]
⇒ sin y = 2x√(1 - x2 )
Differentiating this relationship with respect to x, we obtain
15. Find dy/dx
y = sec-1 [ 1/(2x2 - 1), 0 < x < 1√2
y = sec-1 [ 1/(2x2 - 1), 0 < x < 1√2
Solution
The given relationship is y = sec-1 [ 1/(2x2 - 1)
y = sec-1 [ 1/(2x2 - 1)
⇒ sec y = 1/(2x2 - 1)
⇒ cos y = 2x2 - 1
⇒ 2x2 = 1 + cos y
⇒ 2x2 = 2 cos2 (y/2)
⇒ x = cos y/2
Differentiating this relationship with respect to x, we obtain
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiOWJJXU0iDKa5VnIu3y623DsLb9CW8NLfNW9azU5OFqshBxCJ-p4TTFKIVrshbRy4Hzfx-hY-LnUAtWukXFBZMw_mqoHtl-ENOb3zxW1A_xYJhZL-YDlh1C4RVcGblN6Yv3K9F3qyZLN9y_xcEqSUSMhGsB5ecld-1BAb7cPAPJWEh7Rp9oVRhjE78/w206-h308-rw/Class%2012%20NCERT%20Solution%20Chapter%205%20Continuity%20and%20Differentiability%20Exercise%20-%205.3%20img%2018.JPG)
y = sec-1 [ 1/(2x2 - 1)
⇒ sec y = 1/(2x2 - 1)
⇒ cos y = 2x2 - 1
⇒ 2x2 = 1 + cos y
⇒ 2x2 = 2 cos2 (y/2)
⇒ x = cos y/2
Differentiating this relationship with respect to x, we obtain