Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Finding derivative of Implicit functions
Last updated at May 29, 2023 by Teachoo
Ex 5.3, 7 Find ππ¦/ππ₯ in, sin2 π¦ +cosβ‘ π₯π¦ =π sin2 π¦ +cosβ‘ π₯π¦ =π Differentiating both sides π€.π.π‘.π₯ . (π (sin2 π¦ + cosβ‘ π₯π¦))/ππ₯ = (π (π))/ππ₯ (π (sin2 π¦))/ππ₯ + (π (cosβ‘γ π₯γ π¦))/ππ₯= 0 Calculating Derivative of sin2 π¦ & cos (xπ¦) separately Calculating Derivative of ππππ π (π (sin2 π¦))/ππ₯= (π (sin2 π¦))/ππ₯ Γ ππ¦/ππ¦ =(π (sin2(π¦)))/ππ¦ Γ ππ¦/ππ₯ =2 π ππ π¦ Γ (π(sinβ‘γπ¦)γ)/ππ¦ Γ ππ¦/ππ₯ = 2 sinβ‘γπ¦ cosβ‘γπ¦ Γ γ γ ππ¦/ππ₯ Calculating Derivative of πππ (ππ) (π(cosβ‘γ(π₯π¦))γ)/ππ₯= β sinβ‘(π₯π¦) Γ π/ππ₯(π₯π¦) = βsinβ‘(π₯π¦)Γ(π(π₯)/ππ₯.π¦+π(π¦)/ππ₯.π₯) = βsinβ‘(π₯π¦) . (1.π¦+π₯ ππ¦/ππ₯) = βsinβ‘(π₯π¦) (π¦+π₯ ππ¦/ππ₯) = βsinβ‘(π₯π¦) . π¦βsinβ‘γ(π₯π¦) . π₯γ ππ¦/ππ₯ = βπ¦ sinβ‘γ (π₯π¦)γ β sinβ‘γ(π₯π¦) . π₯γ ππ¦/ππ₯ Now, (π (π ππ2π¦))/ππ₯+(π (cosβ‘γ(π₯π¦)) γ)/ππ₯ = 0 Putting values 2 sinβ‘π¦.cosβ‘γπ¦ .γ ππ¦/ππ₯ + (β π¦ sinβ‘(π₯π¦)βπ₯ sinβ‘(π₯π¦) ππ¦/ππ₯) = 0 2 sinβ‘π¦.cosβ‘γπ¦ .γ ππ¦/ππ₯ β π¦ sinβ‘(π₯π¦) β π₯ sinβ‘(π₯π¦) ππ¦/ππ₯ = 0 2 sinβ‘π¦ cosβ‘γπ¦ γ ππ¦/ππ₯ βπ₯ sinβ‘(π₯π¦) ππ¦/ππ₯ = π¦ sinβ‘(π₯π¦) ππ¦/ππ₯ (2 sinβ‘π¦ cosβ‘π¦ β π₯ sinβ‘(π₯π¦) = π¦ sinβ‘(π₯π¦) ππ¦/ππ₯ = (π¦ sinβ‘γ(π₯π¦)γ)/(2 γsin π¦γβ‘γcosβ‘π¦ β π₯ sinβ‘π₯π¦ γ ) π π/π π = (π πππβ‘γ(ππ)γ)/γπππ ππγβ‘γβ π πππβ‘ππ γ