Chapter 5 Class 12 Continuity and Differentiability

Class 12
Important Questions for exams Class 12

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Ex 5.5, 11 Differentiate the functions in, γ(π₯ πππ β‘π₯ ) γ^π₯ + γ(π₯ π ππβ‘π₯ ) γ^(1/π₯)π¦ = γ(π₯ πππ β‘π₯ ) γ^π₯ + γ(π₯ π ππβ‘π₯ ) γ^(1/π₯) Let π’ = γ(π₯ πππ β‘π₯ ) γ^π₯ , π£ = γ(π₯ π ππβ‘π₯ ) γ^(1/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’ =γ (π₯ πππ β‘π₯ ) γ^π₯ Taking log both sides . logβ‘π’ = logγ (π₯ πππ β‘π₯ ) γ^π₯ logβ‘π’ = π₯ . log (π₯ πππ β‘π₯ ) Differentiating both sides π€.π.π‘.π₯. π(logβ‘π’ )/ππ₯ = (π(π₯ . logβ‘(π₯ cosβ‘π₯ ) ) )/ππ₯ π(logβ‘π’ )/ππ’ . ππ’/ππ₯ = (π(π₯ . logβ‘(π₯ cosβ‘π₯ ) ) )/ππ₯ 1/π’ (ππ’/ππ₯) = (π(π₯ . logβ‘(π₯ cosβ‘π₯ ) ) )/ππ₯ (As πππβ‘(π^π )=π . πππβ‘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‘γ(π₯ πππ  π₯)γ+π₯ (π(πππβ‘(π₯ πππ β‘π₯ ) ) )/ππ₯ 1/π’ (ππ’/ππ₯) = logβ‘γ(π₯ cosβ‘π₯)γ+π₯/(π₯ cosβ‘π₯ ) Γ (π₯ πππ  π₯)^β² 1/π’ (ππ’/ππ₯) = logβ‘γ(π₯ cosβ‘π₯)γ+1/cosβ‘π₯ Γ (1.cosβ‘π₯+π₯(βsinβ‘π₯ )) 1/π’ (ππ’/ππ₯) = logβ‘γ(π₯ cosβ‘π₯)γ+((cosβ‘π₯ β π₯ sinβ‘π₯ ))/cosβ‘π₯ 1/π’ (ππ’/ππ₯) = logβ‘γ(π₯ cosβ‘π₯)γ+cosβ‘π₯/cosβ‘π₯ β(π₯ sinβ‘π₯)/cosβ‘π₯ 1/π’ (ππ’/ππ₯) = logβ‘γ(π₯ cosβ‘π₯)γ+1βπ₯ π‘ππ π₯ ππ’/ππ₯ = u (1βπ₯ tanβ‘π₯+logβ‘γ(π₯ cosβ‘π₯ γ)) Putting value of π’ ππ’/ππ₯ = (π₯ cosβ‘π₯ )^π₯ (1βπ₯ tanβ‘π₯+πππβ‘(π πππβ‘π ) ) Calculating ππ/ππ π£=γ(π₯ π ππβ‘π₯ ) γ^(1/π₯) Taking log both sides logβ‘π£=logβ‘γ γ(π₯ π ππβ‘π₯ ) γ^(1/π₯) γ logβ‘π£= 1/π₯ log (π₯ π ππβ‘π₯ ) Differentiating both sides π€.π.π‘.π₯. (π(logβ‘π£))/ππ₯ = π(1/π₯ " . " )/ππ₯ (π(logβ‘π£))/ππ£ . ππ£/ππ₯ = (1/π₯ .logβ‘(π₯ sinβ‘π₯ ) )^β² 1/π£ Γ ππ£/ππ₯ = (1/π₯ .logβ‘(π₯ sinβ‘π₯ ) )^β² 1/π£ Γ ππ£/ππ₯ = (((logβ‘γ(π₯ sinβ‘π₯)γ )^β² π₯ + π₯^β² logβ‘γ(π₯ sinβ‘π₯)γ)/π₯^2 ) Using quotient rule (π’/π£)^β² = (π’^β² π£ β π£^β² π’)/π£^2 Where u = log (x sin x) , v = x 1/π£ Γ ππ£/ππ₯ = 1/x^2 ((logβ‘γ(π₯ sinβ‘π₯)γ )^β² π₯ + logβ‘γ(π₯ sinβ‘π₯)γ ) 1/π£ Γ ππ£/ππ₯ = 1/x^2 ([1/(π₯ sinβ‘π₯ ) \ Γ(π₯ sinβ‘π₯ )^β² ]π₯ + logβ‘γ(π₯ sinβ‘π₯)γ ) 1/π£ Γ ππ£/ππ₯ = 1/x^2 (1/sinβ‘π₯ \ Γ(π₯ sinβ‘π₯ )^β²+ logβ‘γ(π₯ sinβ‘π₯)γ ) 1/π£ Γ ππ£/ππ₯ = 1/π₯^2 (((1 . π ππβ‘π₯ + π₯ πππ β‘π₯))/π ππβ‘π₯ \ + πππβ‘γ(π₯ π ππβ‘π₯)γ ) 1/π£ Γ ππ£/ππ₯ = 1/π₯^2 (((π ππβ‘π₯ + π₯ πππ β‘π₯))/π ππβ‘π₯ \ + πππβ‘γ(π₯ π ππβ‘π₯)γ ) 1/π£ Γ ππ£/ππ₯ = 1/π₯^2 (1+π₯ cotβ‘π₯+ πππβ‘γ(π₯ π ππβ‘π₯)γ ) ππ£/ππ₯ = π£((1 + γπ₯ πππ‘γβ‘γπ₯ γ β πππβ‘γ (π₯ π ππβ‘π₯ ) γ)/π₯^2 ) ππ£/ππ₯ = (π₯ π ππβ‘π₯ )^(1/π₯) ((1 + γπ₯ πππ‘γβ‘γπ₯ γβ πππ (π₯ π ππβ‘π₯ ))/π₯^2 ) 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ππβ‘π )^π (π β π πππβ‘π+πππβ‘(π πππβ‘π ) ) + (π πππβ‘π )^(π/π) ((γπ πππγβ‘γπ γ + π β π₯π¨π  (π πππβ‘π ))/π^π )