Misc 13 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Jan. 3, 2020 by Teachoo
Transcript
Misc 13 Find ππ¦/ππ₯ , if π¦=γπ ππγ^(βπ) π₯+γπ ππγ^(β1) β(1βπ₯2), β 1 β€ π₯ β€ 1 π¦=γπ ππγ^(βπ) π₯+γπ ππγ^(β1) β(1βπ₯^2 ) , β 1 β€ π₯ β€ 1 Put π₯ = π ππβ‘π π¦=γπ ππγ^(βπ) (sinβ‘π)+γπ ππγ^(β1) β(1βsin^2 π ) π¦=π+γπ ππγ^(β1) β(cos^2 π ) π¦=π+γπ ππγ^(β1) (cos π) π¦=π+γπ ππγ^(β1) (sinβ‘(π/2 βπ) ) π¦=π+ (π/2 βπ) ("As " γπ ππγ^(β1) (sinβ‘γΞΈ)γ=ΞΈ) (As cos ΞΈ = sin (π/2 β ΞΈ)) ("As " γπ ππγ^(β1) (sinβ‘γΞΈ)γ=ΞΈ) π¦=πβπ + π/2 π¦= π/2 Differentiating π€.π.π‘.π₯. ππ¦/ππ₯ = π(π/2)/ππ₯ π π/π π = 0 As derivative of constant is zero, here π/2 is a constant
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.