CBSE Class 12 Sample Paper for 2021 Boards

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

## If y = π π₯ π ππ 2 π₯ + (sin π₯) π₯ , find dy/dx .

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Question 30 If π¦=π^(π₯ γπ ππγ^2β‘π₯ )+(π ππβ‘π₯ )^π₯, find ππ¦/ππ₯ . Let π¦=π^(π₯ γπ ππγ^2β‘π₯ )+(sinβ‘π₯ )^π₯ Let π’ = π^(π₯ γπ ππγ^2β‘π₯ ) & π£ =π^(π₯ γπ ππγ^2β‘π₯ ) π¦ = π’ + π£ Differentiating both sides π€.π.π‘.π₯. ππ¦/ππ₯ = (π (π’ + π£))/ππ₯ ππ¦/ππ₯ = ππ’/ππ₯ + ππ£/ππ₯ Calculating ππ/ππ π’ =π^(π₯ γπ ππγ^2β‘π₯ ) Taking log both sides logβ‘π’=logβ‘γπ^(π₯ γπ ππγ^2β‘π₯ ) γ logβ‘π’=π₯ sin^2β‘π₯.logβ‘π πππβ‘π=π γπππγ^πβ‘π Differentiating both sides π€.π.π‘.π₯. (π(logβ‘π’))/ππ₯ = π(π₯ . sin^2β‘π₯ )/ππ₯ π(logβ‘π’ )/ππ₯ . ππ’/ππ’ = π(π₯ . sin^2β‘π₯ )/ππ₯ π(logβ‘π’ )/ππ’ . ππ’/ππ₯ = π(π₯ . sin^2β‘π₯ )/ππ₯ 1/π’ . ππ’/ππ₯ = π(π₯ . sin^2β‘π₯ )/ππ₯ Using product Rule As (π’π£)β = π’βπ£ + π£βπ’ 1/π’ . ππ’/ππ₯ = ππ₯/ππ₯ . sin^2β‘π₯+ π(sin^2β‘π₯ )/ππ₯ Γ π₯ 1/π’ . ππ’/ππ₯ = 1 . sin^2β‘π₯ + (2 sinβ‘π₯ cosβ‘π₯ ) π₯ 1/π’ . ππ’/ππ₯ = sin^2β‘π₯+π₯ sinβ‘2π₯ ππ’/ππ₯ = u(sin^2β‘π₯+π₯ sinβ‘2π₯ ) ππ/ππ = π^(π γπππγ^πβ‘π ) (γπππγ^πβ‘π+π πππβ‘ππ ) Calculating ππ/ππ π£ =(sinβ‘π₯ )^π₯ Taking log both sides logβ‘π£=logβ‘γ(sinβ‘π₯ )^π₯ γ πππβ‘π=π .π₯π¨π β‘π¬π’π§β‘γπ γ Differentiating both sides π€.π.π‘.π₯. (π(logβ‘π£))/ππ₯ = π(π₯.γ logγβ‘sinβ‘π₯ )/ππ₯ π(logβ‘π£ )/ππ₯ . ππ£/ππ£ = π(π₯ .γ logγβ‘sinβ‘π₯ )/ππ₯ π(logβ‘π£ )/ππ£ . ππ£/ππ₯ = π(π₯ .γ logγβ‘sinβ‘π₯ )/ππ₯ 1/π£ . ππ£/ππ₯ = π(π₯ .γ logγβ‘sinβ‘π₯ )/ππ₯ Using product Rule As (π’π£)β = π’βπ£ + π£βπ’ 1/π£ . ππ£/ππ₯ = ππ₯/ππ₯ . logβ‘sinβ‘π₯ + π(logβ‘sinβ‘π₯ )/ππ₯ Γ π₯ 1/π£ . ππ£/ππ₯ = 1 . logβ‘sinβ‘π₯ + (1/sinβ‘π₯ . π(sinβ‘π₯ )/ππ₯) π₯ 1/π£ . ππ£/ππ₯ = logβ‘sinβ‘π₯ + (1/sinβ‘π₯ . cosβ‘π₯ )π₯ 1/π£ . ππ£/ππ₯ = logβ‘sinβ‘π₯ + π₯ (cosβ‘π₯/sinβ‘π₯ ) ππ£/ππ₯ = v(logβ‘sinβ‘γπ₯+π₯ cotβ‘π₯ γ ) ππ/ππ = (π¬π’π§ π)^π (πππβ‘πππβ‘γπ+π πππβ‘π γ ) Now , ππ¦/ππ₯ = ππ’/ππ₯ + ππ£/ππ₯ Putting value of ππ’/ππ₯ & ππ£/ππ₯ ππ/ππ = π^(π₯ γπ ππγ^2β‘π₯ ) (sin^2β‘π₯+π₯ sinβ‘2π₯ )+(π¬π’π§β‘π )^π (π ππ¨π­β‘π+π₯π¨π β‘π¬π’π§β‘π )