Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Logarithmic Differentiation - Type 1
Example 29 Important
Ex 5.5, 3 Important
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Ex 5.5, 1 Important
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Example 27 Important
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Misc 22 Important
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Example 41 You are here
Misc 9 Important
Example 40 (i)
Logarithmic Differentiation - Type 1
Last updated at June 2, 2023 by Teachoo
Example 41 Find f β²(x) if π (π₯) = (sinβ‘π₯ )^(sin π₯)β‘ for all 0 < x < Ο.Let γπ¦=(sinβ‘π₯ )γ^(sin π₯) Taking log on both sides log π¦=log (γsinβ‘π₯γ^sinβ‘π₯ ) π₯π¨π π=πππβ‘π . π₯π¨π (πππβ‘π ) Differentiating both sides π€.π.π‘. x π(logβ‘π¦)/ππ₯ = π(sinβ‘π₯ . log (sinβ‘π₯ ))/ππ₯ π(logβ‘π¦ )/ππ₯ (ππ¦/ππ¦) = π(sinβ‘π₯ . log (sinβ‘π₯ ))/ππ₯ (As πππβ‘(π^π )=π . πππβ‘π) π(logβ‘π¦ )/ππ¦ (ππ¦/ππ₯) = π(sinβ‘π₯ . log (sinβ‘π₯ ))/ππ₯ 1/π¦ . ππ¦/ππ₯ = π(sinβ‘π₯ . log (sinβ‘π₯ ))/ππ₯ 1/π¦ ππ¦/ππ₯ = π(sinβ‘π₯ )/ππ₯ . γ log γβ‘(sinβ‘π₯ ) + π(γlog γβ‘(sinβ‘π₯ ) )/ππ₯ . sinβ‘π₯ 1/π¦ ππ¦/ππ₯ = cosβ‘π₯ . γlog γβ‘(sinβ‘π₯ ) + 1/sinβ‘π₯ . π(sinβ‘π₯ )/ππ₯ . sinβ‘π₯ 1/π¦ ππ¦/ππ₯ = γcos γβ‘. γlog γβ‘(sinβ‘π₯ ) + 1/sinβ‘π₯ . cosβ‘π₯.sin π₯ 1/π¦ ππ¦/ππ₯ = cosβ‘π₯.log sin π₯+cosβ‘π₯ Using Product rule (uv)β = uβv + vβu where u = sin x & v = log (sin x) ππ¦/ππ₯ = π¦ (cosβ‘π₯.log sin π₯+cosβ‘π₯ ) Putting value of y = (π ππβ‘π₯ )^π ππβ‘π₯ ππ¦/ππ₯ = (sinβ‘π₯ )^sinβ‘π₯ (cosβ‘π₯.log sin π₯+cosβ‘π₯ ) ππ¦/ππ₯ = (sinβ‘π₯ )^sinβ‘π₯ . cosβ‘π₯ (log sin π₯+1) π π/π π = γ(π+π₯π¨π (π¬π’π§ π)) (πππβ‘π )γ^πππβ‘π . πππβ‘π