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Last updated at April 16, 2024 by Teachoo
Example 23 Find ππ¦/ππ₯ , if y + sin y = cosβ‘π₯ y + sin y = cos x Differentiating both sides by x ππ¦/ππ₯ + (π(sinβ‘γπ¦)γ)/ππ₯ = (π (ππ¨π¬β‘γπ)γ)/π π ππ¦/ππ₯ + (π(γsin γβ‘γπ¦)γ)/ππ₯ = β sin x ππ¦/ππ₯ + (π (π¬π’π§β‘γπ)γ)/π π . ππ¦/ππ₯ = β sin x ππ¦/ππ₯ + cos y ππ¦/ππ₯ = β sin x ππ¦/ππ₯ (1 + cos y) = β sin x π π/π π = (βπππβ‘π)/((π + πππβ‘γπ)γ )