Check Full Chapter Explained - Continuity and Differentiability - https://you.tube/Chapter-5-Class-12-Continuity

Last updated at Jan. 3, 2020 by Teachoo

Check Full Chapter Explained - Continuity and Differentiability - https://you.tube/Chapter-5-Class-12-Continuity

Transcript

Ex 5.7, 13 If π¦=3 cosβ‘γ (logβ‘γπ₯)+4 γ sinγβ‘γ (logβ‘γπ₯ )γ γ γ γ, show that π₯2 π¦2 + π₯π¦1 + π¦ = 0 π¦=3 cosβ‘γ (logβ‘γπ₯)+4 γ sinγβ‘γ (logβ‘γπ₯)γ γ γ γ Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = β3 sin (log x) Γ 1/π₯ + 4 cos (log x) Γ 1/π₯ π₯ ππ¦/ππ₯ = β3 sin (log x) + 4 cos (log x) Differentiating π€.π.π‘.π₯ (π₯ ππ¦/ππ₯)^β²= (β3 sin (log x))β + (4 cos (log x))β (π₯ ππ¦/ππ₯)^β²= β3 cos (log x) Γ (log x)β + 4 (βsin (log x)) Γ (log x)β (π₯ ππ¦/ππ₯)^β²= β3 cos (log x) Γ 1/π₯ β 4 sin (log x) Γ 1/π₯ π₯^β² ππ¦/ππ₯ + x (ππ¦/ππ₯)^β²= β3 cos (log x) Γ 1/π₯ β 4 sin (log x) Γ 1/π₯ ππ¦/ππ₯ + π₯ (π^2 π¦)/(ππ₯^2 ) = β3 cos (log x) Γ 1/π₯ β 4 sin (log x) Γ 1/π₯ ππ¦/ππ₯ "+" π₯ (π^2 π¦)/(ππ₯^2 ) = (β1)/π₯ (3 cos (log x) + 4 sin (log x)) ππ¦/ππ₯ + π₯ (π^2 π¦)/(ππ₯^2 ) = (β1)/π₯ Γ y As y = 3 cos (log x) + 4 sin (log x) π₯ (ππ¦/ππ₯+π₯ (π^2 π¦)/(ππ₯^2 )) = βy π₯ ππ¦/ππ₯ + π₯^2 (π^2 π¦)/(ππ₯^2 ) = βy π^π (π ^π π)/(π π^π ) + π π π/π π + y = 0

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.