Example 43 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Dec. 16, 2024 by Teachoo
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Example 43 Differentiate γπ ππγ^2 π₯ π€.π.π‘. π^(cosβ‘π₯ ". " )Let π’ = γπ ππγ^2 π₯ & π£ =π^(cosβ‘π₯ ) We need to differentiate π’ π€.π.π‘. π£ . i.e., ππ’/ππ£ Here, π π/π π = (π π/π π)/(π π/π π) Calculating π π/π π π’ = γπ ππγ^2 π₯ Differentiating π€.π.π‘.π₯. ππ’/ππ₯ = π(γπ ππγ^2 π₯)/ππ₯ ππ’/ππ₯ = 2 sinβ‘π₯ . π(sinβ‘π₯ )/ππ₯ ππ’/ππ₯ = π πππβ‘π . ππ¨π¬β‘π Calculating π π/π π π£ =π^(cosβ‘π₯ ) Differentiating π€.π.π‘.π₯. ππ£/ππ₯ = π(π^(cosβ‘π₯ ) )/ππ₯ ππ£/ππ₯ = π^(cosβ‘π₯ ) . π(cosβ‘π₯ )/ππ₯ ππ£/ππ₯ = π^(cosβ‘π₯ ) . (βsinβ‘π₯ ) ππ£/ππ₯ = βπππβ‘π. π^(πππβ‘π )Therefore ππ’/ππ£ = (ππ’/ππ₯)/(ππ£/ππ₯) = (2 sinβ‘π₯" ." cosβ‘π₯)/(βsinβ‘π₯ . π^(cosβ‘π₯ ) ) = (βπ"." πππβ‘π)/π^(πππβ‘π )
