Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12

Last updated at Jan. 3, 2020 by Teachoo

Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12

Transcript

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)

Miscellaneous

Misc 1

Misc 2

Misc 3

Misc 4

Misc 5 Important

Misc 6 Important

Misc 7 Important

Misc 8

Misc 9 Important

Misc 10

Misc 11 Important

Misc 12

Misc 13 Important

Misc 14 Important

Misc 15 Important

Misc 16 Important

Misc 17 Important You are here

Misc 18

Misc 19 Important

Misc 20

Misc 21

Misc 22

Misc 23 Important

About the Author

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.