This question is similar to Chapter 5 Class 12 Continuity and Differentiability - Ex 5.5
Please check the question here


CBSE Class 12 Sample Paper for 2025 Boards
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
Please check the question here
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 ]