# Misc 13 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at Jan. 3, 2020 by Teachoo

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

Miscellaneous

Misc 1

Misc 2

Misc 3

Misc 4

Misc 5 Important

Misc 6 Important

Misc 7 Important

Misc 8

Misc 9 Important

Misc 10

Misc 11 Important

Misc 12

Misc 13 Important You are here

Misc 14 Important

Misc 15 Important

Misc 16 Important

Misc 17 Important

Misc 18

Misc 19 Important

Misc 20

Misc 21

Misc 22

Misc 23 Important

About the Author

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.