Example 29 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Example 29 Differentiate π₯^sinβ‘π₯ , π₯ > 0 π€.π.π‘. π₯.Let y = π₯^sinβ‘π₯ Taking log both sides logβ‘π¦ = log π₯^sinβ‘π₯ πππβ‘π = πππβ‘π . πππ π Differentiating π€.π.π‘.π₯ (π(logβ‘γπ¦)γ)/ππ₯ = π/ππ₯ (sinβ‘γπ₯ logβ‘π₯ γ ) By product Rule (uv)β = uβv + vβu where u = sin x & v = log x (π(logβ‘γπ¦)γ)/ππ₯ = (π(sinβ‘π₯))/ππ₯.log π₯+sin π₯ . (π(logβ‘π₯))/ππ₯ (π(logβ‘γπ¦)γ)/ππ¦ Γ ππ¦/ππ₯ = cosβ‘π₯ logβ‘π₯ + sinβ‘π₯ 1/π₯ ππ¦/ππ₯ 1/π¦ = πππ πβ‘πππβ‘π + πππβ‘π π/π ππ¦/ππ₯ = π¦ (γπππ π₯γβ‘γπππβ‘γπ₯+1/π₯γ π ππβ‘π₯ γ ) Putting back π¦ = π₯^π ππβ‘π₯ ππ¦/ππ₯ = π₯^π ππβ‘π₯ (cosβ‘γlogβ‘γπ₯+ 1/π₯γ sinβ‘π₯ γ ) = π₯^π ππβ‘π₯ cosβ‘logβ‘π₯ + π₯^π ππβ‘π₯ 1/π₯ π ππβ‘π₯ = π₯^π ππβ‘π₯ cosβ‘logβ‘π₯ + π₯^π ππβ‘π₯ π₯^(β1) sinβ‘π₯ = π^πππβ‘π πππβ‘πππβ‘π + π^πππβ‘γπ β πγ πππβ‘π
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About the Author
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