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Last updated at March 11, 2021 by Teachoo

Ex 5.3, 3 Find ππ¦/ππ₯ in, ππ₯ + ππ¦2 = cosβ‘π¦ ππ₯ + ππ¦2 = cosβ‘π¦ Differentiating both sides π€.π.π‘.π₯ π(ππ₯ + ππ¦2)/ππ₯ = (π (cππ β‘π¦ ))/ππ₯ π(ππ₯)/ππ₯ +π(ππ¦2)/ππ₯= (π γ(cosγβ‘π¦))/ππ₯ a ππ₯/ππ₯ +π π(π¦2)/ππ₯= (π )/ππ₯ cos y a + b (π (π¦2))/ππ₯ Γ ππ¦/ππ¦= (π (cosβ‘π¦ ))/ππ₯ Γππ¦/ππ¦ a + b (π (π¦2))/ππ¦ Γ ππ¦/ππ₯= (π (cosβ‘π¦ ))/ππ¦ Γππ¦/ππ₯ π + π .2π¦Γ ππ¦/ππ₯= βsinβ‘γπ¦ γ ππ¦/ππ₯ π + 2ππ¦ . ππ¦/ππ₯= βsinβ‘γπ¦ γ ππ¦/ππ₯ 2ππ¦ . ππ¦/ππ₯ + sinβ‘γπ¦ γ ππ¦/ππ₯ = 0 β π ππ¦/ππ₯ (2ππ¦ + sinβ‘γπ¦ γ) = βπ π π/π π = (βπ)/(πππ " + " πππβ‘γπ γ )