Derivatives in parametric form

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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### Transcript

Example 41 Differentiate γπ ππγ^2 π₯ π€.π.π‘. π^(cosβ‘π₯ ". " )Let π’ = γπ ππγ^2 π₯ & π£ =π^(cosβ‘π₯ ) We need to differentiate π’ π€.π.π‘. π£ . i.e., ππ’/ππ£ Here, ππ/ππ = (ππ/ππ)/(ππ/ππ) Calculating ππ/ππ π’ = γπ ππγ^2 π₯ Differentiating π€.π.π‘.π₯. ππ’/ππ₯ = π(γπ ππγ^2 π₯)/ππ₯ ππ’/ππ₯ = 2 sinβ‘π₯ . π(sinβ‘π₯ )/ππ₯ ππ’/ππ₯ = π πππβ‘π . ππ¨π¬β‘π Calculating ππ/ππ π£ =π^(cosβ‘π₯ ) Differentiating π€.π.π‘.π₯. ππ£/ππ₯ = π(π^(cosβ‘π₯ ) )/ππ₯ ππ£/ππ₯ = π^(cosβ‘π₯ ) . π(cosβ‘π₯ )/ππ₯ ππ£/ππ₯ = π^(cosβ‘π₯ ) . (βsinβ‘π₯ ) ππ£/ππ₯ = βπππβ‘π. π^(πππβ‘π ) Therefore ππ’/ππ£ = (ππ’/ππ₯)/(ππ£/ππ₯) = (2 sinβ‘π₯" ." cosβ‘π₯)/(βsinβ‘π₯ . π^(cosβ‘π₯ ) ) = (βπ"." πππβ‘π)/π^(πππβ‘π )

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.