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

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

  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Serial order wise

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’™

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.