CBSE Class 12 Sample Paper for 2025 Boards
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CBSE Class 12 Sample Paper for 2025 Boards
Last updated at Feb. 13, 2025 by Teachoo
This question is similar to Chapter 5 Class 12 Continuity and Differentiability - Ex 5.5
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![Question 23 (B) - Differentiate (cos x)^x with respect to x [Video] - CBSE Class 12 Sample Paper for 2025 Boards](https://cdn.teachoo.com/73d2388b-b7ce-4642-9486-b915cce781e4/slide66.jpg)
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Transcript
Question 23 (B) Differentiate the following function with respect to x:(cos π₯)^π₯; (where β π₯β(0,π/2)).Let π¦ = γ(πππ β‘π₯ ) γ^π₯ We use log differentiation Taking log both sides . logβ‘π¦ = logγ (πππ β‘π₯ ) γ^π₯ πππβ‘π = π . πππ (πππβ‘π ) Differentiating both sides π€.π.π‘.π₯. (As πππβ‘(π^π )=π . πππβ‘π) π(logβ‘π¦ )/ππ₯ = (π(π₯ . logβ‘(cosβ‘π₯ ) ) )/ππ₯ π(logβ‘π¦ )/ππ¦ . ππ¦/ππ₯ = (π(π₯ . logβ‘(cosβ‘π₯ ) ) )/ππ₯ 1/π¦ (ππ¦/ππ₯) = (π(π₯ . logβ‘(cosβ‘π₯ ) ) )/ππ₯ 1/π¦ (ππ¦/ππ₯) = ππ₯/ππ₯ logβ‘γ( πππ π₯)γ+π₯ (π(πππβ‘(πππ β‘π₯ ) ) )/ππ₯ 1/π¦ (ππ¦/ππ₯) = logβ‘γ(cosβ‘π₯)γ+π₯ Γ 1/cosβ‘π₯ Γ (πππ π₯)^β² 1/π¦ (ππ¦/ππ₯) = logβ‘γ(cosβ‘π₯)γ+π₯/cosβ‘π₯ Γ (βsinβ‘π₯ ) Using product Rule As (π’π£)β = π’βπ£ + π£βπ’ 1/π¦ (ππ¦/ππ₯) = logβ‘γ(cosβ‘π₯)γβπ₯ tanβ‘π₯ ππ¦/ππ₯ = π¦[logβ‘(cosβ‘π₯ )βπ₯ tanβ‘π₯ ] Putting value of π¦ π π/π π = (ππ¨π¬β‘π )^π [πππβ‘(πππβ‘π )βπ πππβ‘π ]
