Question 16 (OR 1 st question)

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

If y = x^sin ⁡x + sin⁡ (x^x), find dy/dx - Teachoo - CBSE Class 12 Sam

Question 16 (Or 1st) - CBSE Class 12 Sample Paper for 2019 Boards - Part 2
Question 16 (Or 1st) - CBSE Class 12 Sample Paper for 2019 Boards - Part 3 Question 16 (Or 1st) - CBSE Class 12 Sample Paper for 2019 Boards - Part 4 Question 16 (Or 1st) - CBSE Class 12 Sample Paper for 2019 Boards - Part 5 Question 16 (Or 1st) - CBSE Class 12 Sample Paper for 2019 Boards - Part 6

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Question 16 (OR 1st question) If y = π‘₯^sin⁑π‘₯ +sin⁑〖(π‘₯^π‘₯)γ€—, find 𝑑𝑦/𝑑π‘₯ Let u = π‘₯^sin⁑π‘₯ , 𝑣=sin⁑〖(π‘₯^π‘₯)γ€— Thus, y = u + v Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ Calculating derivative of u and v separately Solving 𝒅𝒖/𝒅𝒙 u = π‘₯^sin⁑π‘₯ Taking log both sides log⁑𝑒 = log π‘₯^sin⁑π‘₯ log⁑𝑒 = sin⁑π‘₯ . log π‘₯ Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ (𝑑(log⁑〖𝑒)γ€—)/𝑑π‘₯ = 𝑑/𝑑π‘₯ (sin⁑〖π‘₯ log⁑π‘₯ γ€— ) By product Rule (uv)’ = u’v + v’u where u = sin x & v = log x (𝑑(log⁑〖𝑒)γ€—)/𝑑π‘₯ = (𝑑(sin⁑π‘₯))/𝑑π‘₯.log π‘₯+sin π‘₯ . (𝑑(log⁑π‘₯))/𝑑π‘₯ (𝑑(log⁑〖𝑒)γ€—)/𝑑𝑒 Γ— 𝑑𝑒/𝑑π‘₯ = cos⁑π‘₯ log⁑π‘₯ + sin⁑π‘₯ 1/π‘₯ 1/𝑒 " Γ— " 𝑑𝑒/𝑑π‘₯ = cos π‘₯⁑log⁑π‘₯ + sin⁑π‘₯ 1/π‘₯ 𝑑𝑒/𝑑π‘₯ = 𝑒 (γ€–cos x〗⁑〖log⁑〖π‘₯+ 1/π‘₯γ€— sin⁑π‘₯ γ€— ) Putting back 𝑒 = π‘₯^𝑠𝑖𝑛⁑π‘₯ 𝑑𝑒/𝑑π‘₯ = π‘₯^𝑠𝑖𝑛⁑π‘₯ (cos⁑〖log⁑〖π‘₯+ 1/π‘₯γ€— sin⁑π‘₯ γ€— ) Solving 𝒅𝒗/𝒅𝒙 v = sin⁑〖(π‘₯^π‘₯)γ€— Let t = π‘₯^π‘₯ Taking log both sides log⁑𝑑 = log π‘₯^π‘₯ log⁑𝑑 = π‘₯ log π‘₯ Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ (𝑑(log⁑〖𝑑)γ€—)/𝑑π‘₯ = 𝑑/𝑑π‘₯ (π‘₯ log⁑π‘₯ ) (𝑑(log⁑〖𝑑)γ€—)/𝑑π‘₯ = (𝑑(π‘₯))/𝑑π‘₯.log π‘₯+π‘₯ . (𝑑(log⁑π‘₯))/𝑑π‘₯ (𝑑(log⁑〖𝑑)γ€—)/𝑑π‘₯ = (𝑑(π‘₯))/𝑑π‘₯.log π‘₯+π‘₯ . (𝑑(log⁑π‘₯))/𝑑π‘₯ (𝑑(log⁑〖𝑑)γ€—)/𝑑𝑑 Γ— 𝑑𝑑/𝑑π‘₯ = 1.log π‘₯+π‘₯Γ—1/π‘₯ 1/t Γ— 𝑑𝑑/𝑑π‘₯ = log π‘₯+1 𝑑𝑑/𝑑π‘₯ = t(log π‘₯+1) Putting t = xx 𝑑𝑑/𝑑π‘₯ = π‘₯^π‘₯ (π‘™π‘œπ‘” π‘₯+1) Now, v = sin⁑〖(π‘₯^π‘₯)γ€— v = sin⁑〖(𝑑)γ€— Differentiating w.r.t. x 𝑑𝑣/𝑑π‘₯ = (𝑑(sin⁑𝑑))/𝑑π‘₯ 𝑑𝑣/𝑑π‘₯ = (𝑑(sin⁑𝑑))/𝑑𝑑 Γ— 𝑑𝑑/𝑑π‘₯ 𝑑𝑣/𝑑π‘₯ = cos t Γ— 𝑑𝑑/𝑑π‘₯ 𝑑𝑣/𝑑π‘₯ = cos⁑〖(π‘₯^π‘₯)γ€— (π‘™π‘œπ‘” π‘₯+1)π‘₯^π‘₯ Now, from (1) 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = π‘₯^𝑠𝑖𝑛⁑π‘₯ (cos⁑〖log⁑〖π‘₯+ 1/π‘₯γ€— sin⁑π‘₯ γ€— )+ cos⁑〖〖(π‘₯γ€—^π‘₯)γ€— (π‘™π‘œπ‘” π‘₯+1)π‘₯^π‘₯

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.