# Ex 5.3, 8 - 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, 8 Find ππ¦/ππ₯ in, sin2 π₯ + cos2 π¦ = 1 sin2 π₯ + cos2 π¦ = 1 Differentiating both sides π€.π.π‘.π₯ . (π (sin2 π₯ + cos2 π¦))/ππ₯ = (π (1))/ππ₯ (π (sin2 π₯))/ππ₯ + (π (cosβ‘2 π¦))/ππ₯ = 0 Calculating Derivative of sin2 π₯ & cos^2 π¦ sepretaly Finding Derivative of ππππ π (π (sin2 π₯))/ππ₯ =2γ π ππγ^(2β1) π₯ . (π(sin^2β‘π₯))/ππ₯ =2 sinβ‘π₯ . (π(sinβ‘π₯))/ππ₯ (Derivative of constant is 0) =2 sinβ‘γπ₯ γ cosγβ‘π₯ γ Finding Derivative of γπππγ^π π (π (cos2 π¦))/ππ₯ =2γcosβ‘π¦γ^(2β1) ". " π/ππ₯ " "(cosβ‘π¦) =2 cosβ‘π¦ . (βsinβ‘π¦) . (π(π¦))/ππ₯ =β 2 cosβ‘π¦ sinβ‘π¦ . ππ¦/ππ₯ Now, (π (sin2 π₯))/ππ₯+ (π (cos2 π¦))/ππ₯ = 0 2 sinβ‘π₯ .cosβ‘π₯ + (β 2 cosβ‘π¦ sinβ‘π¦ ". " ππ¦/ππ₯)= 0 2 sinβ‘π₯ .cosβ‘π₯ β 2 sinβ‘π¦β‘γ .γ cosβ‘π¦ . ππ¦/ππ₯ = 0 β 2 sinβ‘π¦β‘γ .γ cosβ‘π¦ . ππ¦/ππ₯ = β 2 sinβ‘π₯ cosβ‘π₯ β sinβ‘2π¦β‘γ .γ ππ¦/ππ₯ = β sinβ‘2π₯ ππ¦/ππ₯ = γβ sinγβ‘2π₯/(βsinβ‘2π¦ ) π π/π π = πππβ‘ππ/πππβ‘ππ (2 sin x cos x = sin 2x)