# Ex 5.3, 4 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at April 16, 2024 by Teachoo

Last updated at April 16, 2024 by Teachoo

Ex 5.3, 4 Find ππ¦/ππ₯ in, π₯π¦ + π¦2 = tanβ‘π₯ + π¦ π₯π¦ + π¦2 = tanβ‘π₯ + π¦ Differentiating both sides π€.π.π‘.π₯ π(π₯π¦ + π¦2)/ππ₯ = (π (tanβ‘π₯ + π¦" " ))/ππ₯ π(π₯π¦)/ππ₯ +π(π¦2)/ππ₯= (π (tanβ‘π₯))/ππ₯ + (π (π¦))/ππ₯ Using product rule in (π₯π¦)^β²=π₯^β² π¦+π¦^β² π₯ ((π(π₯))/ππ₯. π¦+ (π(π¦))/ππ₯.π₯) + π(π¦2)/ππ₯= (π (tanβ‘π₯))/ππ₯ + ππ¦/ππ₯ 1.π¦ + π₯.ππ¦/ππ₯ + (π(π¦2) )/ππ₯ Γππ¦/ππ¦= sec2 π₯ + ππ¦/ππ₯ π¦ + π₯.ππ¦/ππ₯ + (π(π¦2) )/ππ¦ Γππ¦/ππ₯ = sec2 π₯ + ππ¦/ππ₯ π¦ + π₯.ππ¦/ππ₯ + 2y .ππ¦/ππ₯ =sec2 π₯ + ππ¦/ππ₯ π₯.ππ¦/ππ₯ + 2y .ππ¦/ππ₯ " "β ππ¦/ππ₯ =sec2 π₯βπ¦ ππ¦/ππ₯ (π₯+2π¦β1)=sec2 π₯βπ¦ π π/π π = (ππππ π β π" " )/(π + ππ β π)