# Example 37 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 37 (Method 1) Find ππ¦/ππ₯ , if π₯^(2/3) + π¦^(2/3) = π^(2/3) . π₯^(2/3) + π¦^(2/3) = π^(2/3) Differentiating w.r.t. x (π(π₯^(2/3)))/ππ₯ + (π(π¦^(2/3)))/ππ₯ = (π(π^(2/3)))/ππ₯ 2/3 π₯^(2/3 β 1) + (π(π¦^(2/3)))/ππ₯ Γ ππ¦/ππ¦ = 0 2/3 π₯^((β1)/3) + (π(π¦^(2/3)))/ππ¦ Γ ππ¦/ππ₯ = 0 2/3 π₯^((β1)/3) + 2/3 π¦^(2/3 β 1) Γ ππ¦/ππ₯ = 0 2/3Γ1/π₯^(1/3) + 2/3 y^((β1)/3) Γ ππ¦/ππ₯ = 0 2/3Γ1/π₯^(1/3) + 2/3 Γ 1/π¦^(1/3) Γ ππ¦/ππ₯ = 0 2/3Γ1/π¦^(1/3) Γ ππ¦/ππ₯ = 0 β 2/3Γ 1/π₯^(1/3) 2/3Γ1/π¦^(1/3) Γ ππ¦/ππ₯ = (β2)/3Γ1/π₯^(1/3) 1/π¦^(1/3) Γ ππ¦/ππ₯ = (β1)/π₯^(1/3) ππ¦/ππ₯ = β π¦^(1/3)/π₯^(1/3) π π/π π = β (π/π)^(π/π) Example 37 (Method 2) Find ππ¦/ππ₯ , if π₯^(2/3) + π¦^(2/3) = π^(2/3) . π₯^(2/3) + π¦^(2/3) = π^(2/3) π₯^(2/3) + π¦^(2/3) = π^(2/3) . 1 π₯^(2/3) + π¦^(2/3) = π^(2/3) . (sin^2β‘π+cos^2β‘π ) π₯^(2/3) + π¦^(2/3) = π^(2/3) sin^2β‘π+π^(2/3) cos^2β‘π Let π^(π/π) = π^(π/π) γπππγ^πβ‘π½ Cubing both sides π₯^(2/3 \ Γ 3) = π^(2/3 Γ 3) (cos^2β‘π )^3 π₯^2 = π^2 cos^6β‘π x = π cos3 π Let π^(π/π) = π^(π/π) γπ¬π’π§γ^πβ‘π½ Cubing both sides π¦^(2/3 \ Γ 3) = π^(2/3 Γ 3) (sin^2β‘π )^3 π¦^2 = π^2 sin^6β‘π y = a sin3 π Let π₯ = a cos^3β‘π & π¦ = a sin^3β‘π Putting values of x and y in L.H.S of equation Solving LHS π₯^(2/3) + π¦^(2/3) = (a cos^3β‘π )^(2/3) + (a sin^3β‘π )^(2/3) = a^(2/3) cos^(3 Γ 2/3)β‘π + a^(2/3) sin^(3 Γ 2/3)β‘π = a^(2/3) cos^2β‘π + a^(2/3) sin^2β‘π = a^(2/3) (cos^2β‘π+sin^2β‘π ) = a^(2/3) . 1 = a^(2/3) = R.H.S Hence it is proved that π₯ = a cos^3β‘π & π¦ = a sin^3β‘π Now, ππ¦/ππ₯ = ππ¦/ππ₯ . ππ/ππ = ππ¦/ππ . ππ/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ). Calculating π π/π π½ π¦ = a sin^3β‘π ππ¦/ππ = π(a sin^3β‘π )/ππ₯ ππ¦/ππ = a 3 sin^(3 β 2)β‘π . π(sinβ‘π )/ππ₯ ππ¦/ππ = 3π sin2 π cos π Calculating π π/π π½ π₯ = a cos^3β‘π ππ₯/ππ = (π (γa cosγ^3β‘γπ)γ)/ππ ππ₯/ππ = 3a cos^(3 β1)β‘π (π (πππ π½) )/π π½ ππ₯/ππ = 3a cos^2β‘π . (βsinβ‘π ) ππ₯/ππ = β3a π ππβ‘π cos^2β‘π Therefore ππ¦/ππ₯ = ππ¦/ππ Γ· ππ₯/ππ = (3a sin^2β‘π cosβ‘π)/(β3a cos^2β‘π sinβ‘π ) = (βsinβ‘π)/cosβ‘π = βtan π β΄ ππ¦/ππ₯ = βtan π Finding value of tan π π¦/π₯ = (π sin^3β‘π)/(π cos^3β‘π ) (π sin^3β‘π)/(π cos^3β‘π ) = π¦/π₯ sin^3β‘π/cos^3β‘π = π¦/π₯ (sinβ‘π/cosβ‘π )^3 =π¦/π₯ sinβ‘π/cosβ‘π = (π¦/π₯)^(1/3) tanβ‘π = β(π¦/π₯) Thus, ππ¦/ππ₯ = βtan π π π/π π = ββ(π&π/π)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.