Derivatives in parametric form

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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Ex 5.6, 6 If x and y are connected parametrically by the equations without eliminating the parameter, Find ππ¦/ππ₯, π₯ = π (π β sinβ‘π), π¦ = π (1 + cosβ‘π)Here ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) Calculating ππ/ππ½ π¦ = π (1+cosβ‘π) ππ¦/ππ = π(π (1+cosβ‘π))/ππ ππ¦/ππ = π (π(1 + cosβ‘π )/ππ) ππ¦/ππ = π (0βsinβ‘π ) ππ¦/ππ = π (βsinβ‘π ) ππ/ππ½ = βπ πππβ‘π½ Calculating ππ/ππ½ π₯=π (π βsinβ‘π ) ππ₯/ππ = π(π π βπ sinβ‘π )/ππ ππ₯/ππ = π(π π)/ππ β π(π sinβ‘π )/ππ ππ₯/ππ = πβγπ cosγβ‘π ππ/ππ½ = π(γπβπππγβ‘π½ ) Therefore, ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) ππ¦/ππ₯ = (βπ sinβ‘π)/π(γ1 β cosγβ‘π ) ππ¦/ππ₯ = (βsinβ‘π)/γ1 β cosγβ‘π ππ¦/ππ₯ = (β2 γsin γβ‘γπ/2γ γcos γβ‘γπ/2γ)/(2 γsin^2 γβ‘γπ/2γ ) ππ¦/ππ₯ = (βγcos γβ‘γπ/2γ)/(sin π/2) ππ/ππ = βπππβ‘γπ½/πγ Rough We know that sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ Replacing ΞΈ by π/2 sin ΞΈ = 2 πππβ‘γπ½/πγ πππβ‘γπ½/πγ and cos 2ΞΈ = 2cos2 ΞΈ β 1 Replacing ΞΈ by π/2 cos ΞΈ = 2cos2 π½/π β 1