CBSE Class 12 Sample Paper for 2019 Boards

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

Question 16 (OR 1 st question)

If y = x sin ⁡x + sin⁡ (x x ), find dy/dx

Transcript

Question 16 (OR 1st question) If y = π₯^sinβ‘π₯ +sinβ‘γ(π₯^π₯)γ, find ππ¦/ππ₯ Let u = π₯^sinβ‘π₯ , π£=sinβ‘γ(π₯^π₯)γ Thus, y = u + v Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = ππ’/ππ₯ + ππ£/ππ₯ Calculating derivative of u and v separately Solving ππ/ππ u = π₯^sinβ‘π₯ Taking log both sides logβ‘π’ = log π₯^sinβ‘π₯ logβ‘π’ = sinβ‘π₯ . log π₯ Differentiating π€.π.π‘.π₯ (π(logβ‘γπ’)γ)/ππ₯ = π/ππ₯ (sinβ‘γπ₯ logβ‘π₯ γ ) By product Rule (uv)β = uβv + vβu where u = sin x & v = log x (π(logβ‘γπ’)γ)/ππ₯ = (π(sinβ‘π₯))/ππ₯.log π₯+sin π₯ . (π(logβ‘π₯))/ππ₯ (π(logβ‘γπ’)γ)/ππ’ Γ ππ’/ππ₯ = cosβ‘π₯ logβ‘π₯ + sinβ‘π₯ 1/π₯ 1/π’ " Γ " ππ’/ππ₯ = cos π₯β‘logβ‘π₯ + sinβ‘π₯ 1/π₯ ππ’/ππ₯ = π’ (γcos xγβ‘γlogβ‘γπ₯+ 1/π₯γ sinβ‘π₯ γ ) Putting back π’ = π₯^π ππβ‘π₯ ππ’/ππ₯ = π₯^π ππβ‘π₯ (cosβ‘γlogβ‘γπ₯+ 1/π₯γ sinβ‘π₯ γ ) Solving ππ/ππ v = sinβ‘γ(π₯^π₯)γ Let t = π₯^π₯ Taking log both sides logβ‘π‘ = log π₯^π₯ logβ‘π‘ = π₯ log π₯ Differentiating π€.π.π‘.π₯ (π(logβ‘γπ‘)γ)/ππ₯ = π/ππ₯ (π₯ logβ‘π₯ ) (π(logβ‘γπ‘)γ)/ππ₯ = (π(π₯))/ππ₯.log π₯+π₯ . (π(logβ‘π₯))/ππ₯ (π(logβ‘γπ‘)γ)/ππ₯ = (π(π₯))/ππ₯.log π₯+π₯ . (π(logβ‘π₯))/ππ₯ (π(logβ‘γπ‘)γ)/ππ‘ Γ ππ‘/ππ₯ = 1.log π₯+π₯Γ1/π₯ 1/t Γ ππ‘/ππ₯ = log π₯+1 ππ‘/ππ₯ = t(log π₯+1) Putting t = xx ππ‘/ππ₯ = π₯^π₯ (πππ π₯+1) Now, v = sinβ‘γ(π₯^π₯)γ v = sinβ‘γ(π‘)γ Differentiating w.r.t. x ππ£/ππ₯ = (π(sinβ‘π‘))/ππ₯ ππ£/ππ₯ = (π(sinβ‘π‘))/ππ‘ Γ ππ‘/ππ₯ ππ£/ππ₯ = cos t Γ ππ‘/ππ₯ ππ£/ππ₯ = cosβ‘γ(π₯^π₯)γ (πππ π₯+1)π₯^π₯ Now, from (1) ππ¦/ππ₯ = ππ’/ππ₯ + ππ£/ππ₯ ππ¦/ππ₯ = π₯^π ππβ‘π₯ (cosβ‘γlogβ‘γπ₯+ 1/π₯γ sinβ‘π₯ γ )+ cosβ‘γγ(π₯γ^π₯)γ (πππ π₯+1)π₯^π₯