Ex 5.3, 3 - Find dy/dx in, ax+by2 = cos y - Chapter 5 NCERT

Ex 5.3, 3 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

Take a fresh quiz. Then take another.
Every attempt is a new AI-adaptive Teachoo quiz with 3 questions, selected from your answers, mistakes, and progress.
Remove Ads

Transcript

Ex 5.3, 3 Find 𝑑𝑦/𝑑π‘₯ in, π‘Žπ‘₯ + 𝑏𝑦2 = cos⁑𝑦 π‘Žπ‘₯ + 𝑏𝑦2 = cos⁑𝑦 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑(π‘Žπ‘₯ + 𝑏𝑦2)/𝑑π‘₯ = (𝑑 (cπ‘œπ‘ β‘π‘¦ ))/𝑑π‘₯ 𝑑(π‘Žπ‘₯)/𝑑π‘₯ +𝑑(𝑏𝑦2)/𝑑π‘₯= (𝑑 γ€–(cos〗⁑𝑦))/𝑑π‘₯ a 𝑑π‘₯/𝑑π‘₯ +𝑏 𝑑(𝑦2)/𝑑π‘₯= (𝑑 )/𝑑π‘₯ cos y a + b (𝑑 (𝑦2))/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦= (𝑑 (cos⁑𝑦 ))/𝑑π‘₯ ×𝑑𝑦/𝑑𝑦 a + b (𝑑 (𝑦2))/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯= (𝑑 (cos⁑𝑦 ))/𝑑𝑦 ×𝑑𝑦/𝑑π‘₯ π‘Ž + 𝑏 .2𝑦× 𝑑𝑦/𝑑π‘₯= βˆ’sin⁑〖𝑦 γ€— 𝑑𝑦/𝑑π‘₯ π‘Ž + 2𝑏𝑦 . 𝑑𝑦/𝑑π‘₯= βˆ’sin⁑〖𝑦 γ€— 𝑑𝑦/𝑑π‘₯ 2𝑏𝑦 . 𝑑𝑦/𝑑π‘₯ + sin⁑〖𝑦 γ€— 𝑑𝑦/𝑑π‘₯ = 0 βˆ’ π‘Ž 𝑑𝑦/𝑑π‘₯ (2𝑏𝑦 + sin⁑〖𝑦 γ€—) = βˆ’π‘Ž π’…π’š/𝒅𝒙 = (βˆ’π’‚)/(πŸπ’ƒπ’š " + " π’”π’Šπ’β‘γ€–π’š γ€— )

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh is an IIT Kanpur graduate and has been teaching for 16+ years. At Teachoo, he breaks down Maths, Science and Computer Science into simple steps so students understand concepts deeply and score with confidence.

Many students prefer Teachoo Black for a smooth, ad-free learning experience.