# Misc 17 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at Jan. 3, 2020 by Teachoo

Last updated at Jan. 3, 2020 by Teachoo

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Misc 17 If π₯=π (cosβ‘π‘ + π‘ sinβ‘π‘) and y=π (sinβ‘π‘ β π‘ cosβ‘π‘), Find (π^2 π¦)/γππ₯γ^2 If π₯=π (cosβ‘π‘ + π‘ sinβ‘π‘)\ & π¦=π (sinβ‘π‘ β π‘ cosβ‘π‘) We need to find (π^2 π¦)/γππ₯γ^2 First we find ππ¦/ππ₯ ππ¦/ππ₯ = ππ¦/ππ₯ . ππ‘/ππ‘ ππ¦/ππ₯ = ππ¦/ππ‘ . ππ‘/ππ₯ ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) Calculating π π/π π π¦=π (sinβ‘π‘β π‘ cosβ‘π‘ ) Differentiating π€.π.π‘.π₯. ππ¦/ππ‘ = π(π (sinβ‘π‘β π‘ cosβ‘π‘ ))/ππ‘ ππ¦/ππ‘ = π π(sinβ‘π‘β π‘ cosβ‘π‘ )/ππ‘ ππ¦/ππ‘ = π (π(sinβ‘π‘ )/ππ‘ β π(π‘ cosβ‘π‘ )/ππ‘) ππ¦/ππ‘ = π (cosβ‘π‘β π(π‘ cosβ‘π‘ )/ππ‘) Using product rule As (π’π£)β = π’βπ£ + π£βπ’ where u = t & v = cos t ππ¦/ππ‘ = π (cosβ‘π‘ β(ππ‘/ππ‘ . cosβ‘π‘+ (π cosβ‘π‘)/ππ‘ . π‘ )) ππ¦/ππ‘ = π (cosβ‘π‘ β(cosβ‘π‘+(γβsinγβ‘π‘ ) . π‘)) ππ¦/ππ‘ = π (cosβ‘π‘ β(cosβ‘π‘β(sinβ‘π‘ ) . π‘)) ππ¦/ππ‘ = π (cosβ‘π‘ βcosβ‘π‘+π‘ .sinβ‘π‘ ) ππ¦/ππ‘ = π (0+π‘ sinβ‘π‘ ) ππ¦/ππ‘ = π .π‘.sinβ‘π‘ Hence ππ¦/ππ‘ = π .π‘.sinβ‘π‘ Calculating π π/π π π₯=π (cosβ‘π‘+ π‘ sinβ‘π‘ ) Differentiating π€.π.π‘.π₯. ππ₯/ππ‘ = π(π (cosβ‘π‘ + π‘ sinβ‘π‘)" " )/ππ‘ ππ₯/ππ‘ = π (π(cosβ‘π‘ + π‘ sinβ‘π‘)/ππ‘) ππ₯/ππ‘ = π (π(cosβ‘π‘)/ππ‘ + π(π‘ sinβ‘π‘)/ππ‘) ππ₯/ππ‘ = π (γβsinγβ‘π‘ + π(π‘ sinβ‘π‘ )/ππ‘) Using product rule As (π’π£)β = π’βπ£ + π£βπ’ ππ₯/ππ‘ = π (γβsinγβ‘π‘+(ππ‘/ππ‘ . sinβ‘π‘+ π(sinβ‘π‘ )/ππ‘ . π‘ )) ππ₯/ππ‘ = π (γβsinγβ‘π‘+(sinβ‘π‘+cosβ‘π‘ . π‘)) ππ₯/ππ‘= π (βsinβ‘π‘+sinβ‘π‘+π‘ .cππ β‘π‘ ) ππ₯/ππ‘ = π .π‘.cosβ‘π‘ Now , ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) ππ¦/ππ₯ = (π" " .π‘.sinβ‘π‘)/(π" " .π‘.cosβ‘π‘ ) π π/π π = πππβ‘π Again Differentiating π€.π.π‘.π₯. π/ππ₯ (ππ¦/ππ₯) = π(tanβ‘π‘)/ππ₯ (π^2 π¦)/(ππ₯^2 ) = π(tanβ‘π‘)/ππ₯ (π^2 π¦)/(ππ₯^2 ) = π(tanβ‘π‘)/ππ₯ . ππ‘/ππ‘ (π^2 π¦)/(ππ₯^2 ) =sec^2β‘π‘ . ππ‘/ππ₯ (π^2 π¦)/(ππ₯^2 ) =sec^2β‘π‘ Γ· π π/π π (π^2 π¦)/(ππ₯^2 ) = sec^2β‘π‘ Γ· π.π.ππππ (π^2 π¦)/(ππ₯^2 ) = (sec^2β‘π‘ )/(π" " . π‘.cosβ‘π‘ ) "We have calculated" ππ₯/ππ‘ " = " π" ".π‘.πππ β‘π‘ (π^2 π¦)/(ππ₯^2 ) = (sec^2β‘π‘ )/(π" " . π‘ Γ 1/secβ‘π‘ ) (π^2 π¦)/(ππ₯^2 ) = (sec^3β‘π‘ )/(π" " . π‘) Hence (π ^π π)/(π π^π ) = (γπππγ^πβ‘π )/(π" " . 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 9 years. He provides courses for Maths and Science at Teachoo.