Derivatives in parametric form

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

Β

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

### Transcript

Example 33 Find ππ¦/ππ₯ , if π₯ = π (π+sinβ‘π), π¦ = π (1 β cosβ‘π)Here ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) Calculating ππ/ππ½ π¦ = π (1 β cosβ‘π) ππ¦/ππ = π(π (1 β cosβ‘π))/ππ ππ¦/ππ = π (0 β(βsinβ‘π )) ππ¦/ππ = π (πππβ‘π½ ) Calculating ππ/ππ½ π₯ = π (π+π ππβ‘π) ππ₯/ππ = π(π(π + π ππβ‘π))/ππ ππ₯/ππ = π (ππ/ππ+π(sinβ‘π )/(ππ )) ππ₯/ππ = π (π+ππ¨π¬β‘π½ ) Therefore ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) ππ¦/ππ₯ = π" " (sinβ‘π )/π" " (1 +γ cosγβ‘π ) ππ¦/ππ₯ = sinβ‘π/(1 +γ cosγβ‘π ) ππ¦/ππ₯ = (γπ πππγβ‘γ π½/πγ .γ πππγβ‘γ π½/πγ)/(1 + γπ γπππγ^π γβ‘γπ½/πγ β π) ππ¦/ππ₯ = (sinβ‘γ π/2γ )/γcos γβ‘γπ/2γ ππ/ππ = 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