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Example 29 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€— ) 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo