Get live Maths 1-on-1 Classs - Class 6 to 12
Example 35 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at March 22, 2023 by Teachoo
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Example 35 You are here
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Transcript
Example 35 Find ππ¦/ππ₯ , if π₯ = ππ‘2, π¦ = 2ππ‘ . ππ¦/ππ₯ = ππ¦/ππ₯ Γ ππ‘/ππ‘ ππ¦/ππ₯ = ππ¦/ππ‘ Γ ππ‘/ππ₯ ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) Calculating π π/π π π¦ = 2ππ‘" " ππ¦/ππ‘ " " = π(2ππ‘)/ππ‘ ππ¦/ππ‘ = 2π . π(π‘)/ππ‘ ππ¦/ππ‘ = ππ Calculating π π/π π π₯=ππ‘2 ππ₯/ππ‘ " " = π(ππ‘2)/ππ‘ ππ₯/ππ‘ " " = π . π(π‘2)/ππ‘ " " ππ₯/ππ‘ " " = πππ Therefore ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) ππ¦/ππ₯ = 2π/2aπ‘ π π/π π = π/π
