# Ex 5.6, 9 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at April 16, 2024 by Teachoo

Last updated at April 16, 2024 by Teachoo

Ex 5.6, 9 If x and y are connected parametrically by the equations without eliminating the parameter, Find ππ¦/ππ₯, π₯=π sec β‘π, π¦=π tanβ‘πHere ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) Calculating π π/π π½ ππ¦/ππ = π(π tanβ‘π )/ππ ππ¦/ππ = π π(tanβ‘π )/ππ ππ¦/ππ = π .γπ¬ππγ^πβ‘π½ Calculating π π/π π½ ππ₯/ππ = π(π sec β‘π)/ππ ππ₯/ππ = π π(sec β‘π)/ππ ππ₯/ππ = π (πππβ‘π½.πππβ‘π½ ) Therefore ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) ππ¦/ππ₯ = (π .γ secγ^2β‘π)/(π (secβ‘π.tanβ‘π ) ) ππ¦/ππ₯ = (π secβ‘π)/(π tanβ‘π ) ππ¦/ππ₯ = (π . 1/cosβ‘π )/(π (sinβ‘π/cosβ‘π ) ) ππ¦/ππ₯ = π Γ 1/cosβ‘π Γ cosβ‘π/γa sinγβ‘π ππ¦/ππ₯ = π/π (1/sinβ‘π ) π π/π π = π/π πππππ π½