Check sibling questions

Example 32 - Differentiate xsin x - Chapter 5 Class 12 - Examples

Example 32 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

This video is only available for Teachoo black users

Get Real time Doubt solving from 8pm to 12 am!


Transcript

Example 32 Differentiate π‘₯^sin⁑π‘₯ , π‘₯ > 0 𝑀.π‘Ÿ.𝑑. π‘₯.Let y = π‘₯^sin⁑π‘₯ Taking log both sides log⁑𝑦 = log π‘₯^sin⁑π‘₯ π’π’π’ˆβ‘π’š = π’”π’Šπ’β‘π’™ . π’π’π’ˆ 𝒙 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ (𝑑(log⁑〖𝑦)γ€—)/𝑑π‘₯ = 𝑑/𝑑π‘₯ (sin⁑〖π‘₯ log⁑π‘₯ γ€— ) (π‘™π‘œπ‘”β‘γ€–π‘Ž^𝑏=𝑏 π‘™π‘œπ‘”β‘π‘Ž γ€—) By product Rule (uv)’ = u’v + v’u where u = sin x & v = log x (𝑑(log⁑〖𝑦)γ€—)/𝑑π‘₯ = (𝑑(sin⁑π‘₯))/𝑑π‘₯.log π‘₯+sin π‘₯ . (𝑑(log⁑π‘₯))/𝑑π‘₯ (𝑑(log⁑〖𝑦)γ€—)/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯ = cos⁑π‘₯ log⁑π‘₯ + sin⁑π‘₯ 1/π‘₯ 𝑑𝑦/𝑑π‘₯ 1/𝑦 = 𝒄𝒐𝒔 π’™β‘π’π’π’ˆβ‘π’™ + π’”π’Šπ’β‘π’™ 𝟏/𝒙 𝑑𝑦/𝑑π‘₯ = 𝑦 (γ€–π‘π‘œπ‘  π‘₯γ€—β‘γ€–π‘™π‘œπ‘”β‘γ€–π‘₯+1/π‘₯γ€— 𝑠𝑖𝑛⁑π‘₯ γ€— ) Putting back 𝑦 = π‘₯^𝑠𝑖𝑛⁑π‘₯ 𝑑𝑦/𝑑π‘₯ = π‘₯^𝑠𝑖𝑛⁑π‘₯ (cos⁑〖log⁑〖π‘₯+ 1/π‘₯γ€— sin⁑π‘₯ γ€— ) = π‘₯^𝑠𝑖𝑛⁑π‘₯ cos⁑log⁑π‘₯ + π‘₯^𝑠𝑖𝑛⁑π‘₯ 1/π‘₯ 𝑠𝑖𝑛⁑π‘₯ = π‘₯^𝑠𝑖𝑛⁑π‘₯ cos⁑log⁑π‘₯ + π‘₯^𝑠𝑖𝑛⁑π‘₯ π‘₯^(βˆ’1) sin⁑π‘₯ = 𝒙^π’”π’Šπ’β‘π’™ π’„π’π’”β‘π’π’π’ˆβ‘π’™ + 𝒙^π’”π’Šπ’β‘γ€–π’™ βˆ’ πŸγ€— π’”π’Šπ’β‘π’™

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.