Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Logarithmic Differentiation - Type 1
Example 29 Important You are here
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Example 41
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Example 40 (i)
Logarithmic Differentiation - Type 1
Last updated at June 2, 2023 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