# Ex 5.2, 6 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 5.2, 6 Differentiate the functions with respect to 𝑥 cos𝑥3 . sin2 (𝑥5) Let 𝑦 =cos𝑥3 . sin2 (𝑥5) Let 𝑢 = cos𝑥3 & 𝑣=sin2 (𝑥5) 𝑦 = 𝑢𝑣 We need to find derivative of 𝑦 𝑤.𝑟.𝑡.𝑥 i.e. 𝑦′ = 𝑢𝑣′ 𝑑𝑦𝑑𝑥 = 𝑑 𝑢𝑣𝑑𝑥 𝑑𝑑𝑥 𝑢𝑣= 𝑑 𝑢𝑑𝑥 . 𝑣+ 𝑑 𝑣𝑑𝑥 . 𝑢 Finding 𝒖’ 𝑢=cos𝑥3 Differentiating 𝑑𝑢𝑑𝑥 = 𝑑 cos𝑥3𝑑𝑥 = − sin𝑥3 . 𝑑 𝑥3𝑑𝑥 = − sin 𝑥3. 3 𝑥3 −1 = − sin 𝑥3. 3 𝑥2 = − 3 𝑥2 . sin 𝑥3 Now 𝑑𝑦𝑑𝑥 = 𝑑 𝑢𝑑𝑥 . 𝑣+ 𝑑 𝑣𝑑𝑥 . 𝑢 = − 3 𝑥2 . sin 𝑥3 . sin2 𝑥5+10 𝑥4 . sin 𝑥5 . cos 𝑥5 cos𝑥3 =𝟏𝟎 𝒙𝟒 . 𝒔𝒊𝒏 𝒙𝟓 . 𝒄𝒐𝒔 𝒙𝟓. 𝒄𝒐𝒔 𝒙𝟑− 𝟑 𝒙𝟐. 𝒔𝒊𝒏 𝒙𝟑.𝒔𝒊𝒏𝟐 𝒙𝟓

Finding derivative of a function by chain rule

Chapter 5 Class 12 Continuity and Differentiability

Concept wise

- Checking continuity at a given point
- Checking continuity at any point
- Checking continuity using LHL and RHL
- Algebra of continous functions
- Continuity of composite functions
- Checking if funciton is differentiable
- Finding derivative of a function by chain rule
- Finding derivative of Implicit functions
- Finding derivative of Inverse trigonometric functions
- Finding derivative of Exponential & logarithm functions
- Logarithmic Differentiation - Type 1
- Logarithmic Differentiation - Type 2
- Derivatives in parametric form
- Finding second order derivatives - Normal form
- Finding second order derivatives- Implicit form
- Proofs
- Verify Rolles theorem
- Verify Mean Value Theorem

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.