Question 16 (OR 1 st question)

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

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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€— ) (𝑑(log⁑〖𝑒)γ€—)/𝑑π‘₯ = (𝑑(sin⁑π‘₯))/𝑑π‘₯.log π‘₯+sin π‘₯ . (𝑑(log⁑π‘₯))/𝑑π‘₯ 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‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑒 (γ€–cos x〗⁑〖log⁑〖π‘₯+ 1/π‘₯γ€— sin⁑π‘₯ γ€— ) Putting back 𝑦 = π‘₯^𝑠𝑖𝑛⁑π‘₯ 𝑑𝑦/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‘₯ = cost 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.