Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12   1. Chapter 5 Class 12 Continuity and Differentiability
2. Concept wise
3. Finding derivative of a function by chain rule

Transcript

Ex 5.2, 5 Differentiate the functions with respect to 𝑥 : sin⁡〖 (𝑎𝑥 + 𝑏)〗/cos⁡〖 (𝑐𝑥 + 𝑑)〗 Let 𝑦 = sin⁡〖 (𝑎𝑥 + 𝑏)〗/cos⁡〖 (𝑐𝑥 + 𝑑)〗 Let 𝑢 = sin⁡〖 (𝑎𝑥+𝑏)〗 & 𝑣=cos⁡〖 (𝑐𝑥+𝑑)〗 ∴ 𝒚 = 𝒖/𝒗 We need to find derivative of 𝑦 𝑤.𝑟.𝑡.𝑥 𝑑𝑦/𝑑𝑥 = (𝑢/𝑣)^′ 𝑑𝑦/𝑑𝑥 = (𝑢^′ 𝑣 − 〖 𝑣〗^′ 𝑢)/𝑣^2 Finding 𝒖’ 𝑢=sin⁡〖 (𝑎𝑥+𝑏)〗 Derivative of 𝑢 𝑤.𝑟.𝑡.𝑥 𝑑𝑢/𝑑𝑥 =𝑑(sin⁡〖 (𝑎𝑥+𝑏)〗 )/𝑑𝑥 〖=cos 〗⁡(𝑎𝑥+𝑏) . 𝑑(𝑎𝑥 + 𝑏)/𝑑𝑥 〖=cos 〗⁡(𝑎𝑥+𝑏) (𝑎 +0) 〖=𝐚 𝒄𝒐𝒔 〗⁡(𝒂𝒙+𝒃) Finding 𝒗’ 𝑣=cos⁡〖 (𝑐𝑥+𝑑)〗 Derivative of 𝑣 𝑤.𝑟.𝑡.𝑥 𝑑𝑣/𝑑𝑥 = (𝑑(cos⁡〖 (𝑐𝑥 + 𝑑)〗 )^′ )/𝑑𝑥 〖=−si𝑛 〗⁡(𝑐𝑥+𝑑) . 𝑑(𝑐𝑥 + 𝑑)/𝑑𝑥 〖=−sin 〗⁡(𝑐𝑥+𝑏) (𝑐+0) 〖=−𝐜 𝒔𝒊𝒏 〗⁡(𝒄𝒙+𝒃) Now, 𝒅𝒚/𝒅𝒙 = (𝒖^′ 𝒗 − 〖 𝒗〗^′ 𝒖)/𝒗^𝟐 = (〖a cos 〗⁡〖(𝑎𝑥 + 𝑏) .〖 cos〗⁡〖 (𝑐𝑥 + 𝑑)〗 − (−𝑐 〖sin 〗⁡〖(𝑐𝑥 + 𝑑) 〗 ) 〗 (〖sin 〗⁡〖(𝑎𝑥 + 𝑏)〗 ) )/(cos⁡〖 (𝑐𝑥 + 𝑑)〗 )^2 = (〖a cos 〗⁡〖(𝑎𝑥 + 𝑏) .〖 cos〗⁡(𝑐𝑥 + 𝑑) + 𝑐 . 〖sin 〗⁡(𝑐𝑥 + 𝑑) 〗 sin⁡〖(𝑎𝑥 + 𝑏)〗 )/cos^2⁡(𝑐𝑥 + 𝑑) = (〖a cos 〗⁡〖(𝑎𝑥 + 𝑏) .〖 cos〗⁡〖 (𝑐𝑥 + 𝑑)〗 〗 )/cos^2⁡(𝑐𝑥 + 𝑑) + (𝑐 . 〖sin 〗⁡〖(𝑐𝑥 + 𝑑) 〗. sin⁡〖 (𝑎𝑥 + 𝑏)〗 )/cos^2⁡(𝑐𝑥 + 𝑑) = a cos (𝑎𝑥+ 𝑏) 𝟏/𝒄𝒐𝒔⁡〖 (𝒄𝒙 + 𝒅)〗 + 𝑐 . 〖sin 〗⁡(𝑎𝑥+𝑏) . (〖𝒔𝒊𝒏 〗⁡〖(𝒄𝒙 + 𝒅)〗 )/𝒄𝒐𝒔⁡〖 (𝒄𝒙 + 𝒅)〗 𝟏/𝒄𝒐𝒔⁡〖 (𝒄𝒙 + 𝒅)〗 = 𝒂 𝒄𝒐𝒔 (𝑎𝑥+ 𝑏) .𝒔𝒆𝒄⁡〖(𝑐𝑥+𝑑)〗 + 𝒄 . 〖𝒔𝒊𝒏 〗⁡(𝑎𝑥+𝑏).〖𝒕𝒂𝒏 〗⁡(𝑐𝑥+𝑑).𝒔𝒆𝒄⁡(𝑐𝑥+𝑑)

Finding derivative of a function by chain rule

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.