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Ex 5.4, 1 Differentiate 𝑀.π‘Ÿ.𝑑. π‘₯ in , 𝑒^π‘₯/sin⁑π‘₯ Let π’š = 𝑒^π‘₯/sin⁑π‘₯ Let 𝑒 =𝑒^π‘₯ & 𝑣 = sin⁑π‘₯ ∴ π’š = 𝒖/𝒗 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑(𝑦)/𝑑π‘₯ = 𝑑(𝑒/𝑣)/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = ((𝑒)^β€² 𝑣 βˆ’ 〖𝑣 γ€—^β€² 𝑒)/𝑣^2 𝑑𝑦/𝑑π‘₯ = ((𝑒)^β€² 𝑣 βˆ’ 〖𝑣 γ€—^β€² 𝑒)/𝑣^2 π’…π’š/𝒅𝒙 = (𝒅(𝒆^𝒙 )/𝒅𝒙 . π’”π’Šπ’β‘π’™ βˆ’ (𝒅(π’”π’Šπ’β‘π’™ ) )/𝒅𝒙 . 𝒆^𝒙)/(π’”π’Šπ’β‘π’™ )^𝟐 𝑑𝑦/𝑑π‘₯ = (𝑒^π‘₯ . sin⁑π‘₯ βˆ’ cos⁑π‘₯ . 𝑒^π‘₯)/(sin⁑π‘₯ )^2 𝑑𝑦/𝑑π‘₯ = (𝑒^π‘₯ (sin⁑π‘₯ βˆ’ cos⁑π‘₯ ))/(sin⁑π‘₯ )^2 π’…π’š/𝒅𝒙 = (𝒆^𝒙 (π’”π’Šπ’β‘π’™ βˆ’ 𝒄𝒐𝒔⁑𝒙 ))/γ€–π’”π’Šπ’γ€—^πŸβ‘π’™

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.