Class 12 Maths NCERT Solutions for Chapter 5 Continuity and Differentiability Exercise 5.4
![Class 12 Maths NCERT Solutions for Chapter 5 Continuity and Differentiability Exercise 5.4 Class 12 Maths NCERT Solutions for Chapter 5 Continuity and Differentiability Exercise 5.4](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhGjfy7IpbevPpnlVyBQZad8tXZJqhOIULGDYJZ0AOpVyuzmVM5X5yJPfGPjmSZNiCIN3WeqzjQymEKLM8elBf6pfrLdIioGCLMQEJ6yReydtDncT_hHD4i1h7vAKHigGvdi32cvteBF_Dvhte3p14_Jpw4EWU5F4bElC01WbGdsmlNNKw90kZ-_BgA/w661-h312-rw/cbse-class-12-maths-ncert-solutions-chapter-5-exercise-5-4.jpg)
Continuity and Differentiability Exercise 5.4 Solutions
1. Differentiate the following w.r.t. x :
ex /sin x
Solution
Let y = ex /sin x
By using the quotient rule, we obtain
2. Differentiate the following w.r.t. x : esin-1 x
SolutionLet y = esin-1 x
By using the chain rule, we obtain
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiz-UkXxLzLjv5P5sewn9YKzdSjUvYuGMcNZjkOfn0GemvmQIcNTxASTxpdIuK1HxYzMF-NVPrCkPVQ6Y54aRcn0HVWI63mz1XeuxFqqVumazTe7G5llVlGq3OYDtaUdh2Mk88dPHA5vo8Agty3c9VJezsqfnb1h-1ZvZlJw1iKGqekWPI_2jswuXys/w182-h253-rw/Class%2012%20NCERT%20Solution%20Chapter%205%20Continuity%20and%20Differentiability%20Exercise%20-%205.4%20img%202.JPG)
By using the chain rule, we obtain
3. Differentiate the following w.r.t. x: ex3
Solution
Let y = ex3
By using the chain rule, we obtain
4. Differentiate the following w.r.t. x :
sin(tan-1 e-x )
Solution
Let y = sin(tan-1 e-x )
By using the chain rule, we obtain
5. Differentiate the following w.r.t. x :
log(cos ex )
Solution
Let y = log(cos ex )
By using the chain rule, we obtain
6. Differentiate the following w.r.t. x :
ex + ex2 + ex3
Solution
7. Differentiate the following w.r.t. x :
Solution
Let y = ![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgGRKT9LPJ07wym8tGqXoDOVH6rdaRpbQEGfoXpUoOFPfVJjxcQdG_7bk1JvpbsbEfaSfZ6oDq8NTC14FpYhRS_bNtzcJCpqL618z0BnQIlyBdOKChTZ2wDqROmJdEoJHzLW4ymMvxJXTtweWElcRM_Af_ya-hHqZk2q52PYSf_k9kFRXCQalonF5v8/w44-h28-rw/Class%2012%20NCERT%20Solution%20Chapter%205%20Continuity%20and%20Differentiability%20Exercise%20-%205.4%20img%205.JPG)
y2 = e√x
By differentiating this relationship with respect to x, we obtain
y2 = e√x
y2 = e√x
By differentiating this relationship with respect to x, we obtain
y2 = e√x
8. Differentiate the following w.r.t. x : log (log x), x > 1
Solution
Let y = log (log x)
By using the chain rule, we obtain
By using the chain rule, we obtain
9. Differentiate the following w.r.t. x :
cos x/log x , x > 0
Solution
Let y = cos x/log x
By using the quotient rule, we obtain
By using the quotient rule, we obtain
10. Differentiate the following w.r.t. x :
cos (log x + ex ), x > 0
Solution
Let y = cos (log x + ex )
By using the chain rule, we obtain
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgLRLfv6mAZu1oKqRUI_MQYo1HN6qsHPfEBEcz4pvUqIf7gsjIbhIWH1RiNdyeW6NidgLSNbMKKZX9jhkiNgasFLZiP6xo72VePCkTyeKmSHTSsIxgSdyO5mZRxKagUkFb6NEWbv8L8B_euSvK3xF5cy9ey2Q1s3NA--yUYBR-JP1W-xdyqFGr9CgdF/w306-h210-rw/Class%2012%20NCERT%20Solution%20Chapter%205%20Continuity%20and%20Differentiability%20Exercise%20-%205.4%20img%2010.JPG)
By using the chain rule, we obtain