Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12







Last updated at Jan. 3, 2020 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12
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Misc 6 (Method 1) Differentiate w.r.t. x the function, γπππ‘γ^(β1 ) [(β(1 + sinβ‘π₯ ) + β(1 β sinβ‘π₯ ))/(β(1 + sinβ‘γπ₯ γ ) β β(1 βsinβ‘π₯ )) ] , 0<π₯< π/2 Let π¦= γπππ‘γ^(β1 ) [(β(1 + sinβ‘π₯ ) +β(1 β sinβ‘π₯ ))/(β(1 + sinβ‘γπ₯ γ )ββ(1 β sinβ‘π₯ )) ] Rationalizing the sum π¦= γπππ‘γ^(β1 ) [((β(1 + sinβ‘π₯ ) + β(1 β sinβ‘π₯ )))/((β(1 + sinβ‘γπ₯ γ )β β(1 β sinβ‘π₯ )) ) Γ((β(1 + sinβ‘π₯ ) + β(1 β sinβ‘π₯ )))/((β(1+sinβ‘γπ₯ γ )+ β(1 β sinβ‘π₯ )) )] π¦= γπππ‘γ^(β1 ) [(β(1 + sinβ‘π₯ ) + β(1 β sinβ‘π₯ ))^2/((β(1 + sinβ‘γπ₯ γ )β β(1 βγ sinγβ‘π₯ )) (β(1 + sinβ‘γπ₯ γ )+ β(1 βγ sinγβ‘π₯ )) ) ] = γπππ‘γ^(β1 ) [((β(1 + sinβ‘π₯ ) )^2 + (β(1 β sinβ‘π₯ ) )^2+ 2(β(1 + sinβ‘γπ₯ γ ))(β(1 βγ sinγβ‘π₯ )))/((β(1 + sinβ‘γπ₯ γ )β β(1 β γ sinγβ‘π₯ )) (β(1 + sinβ‘γπ₯ γ )+ β(1 β sinβ‘π₯ )) ) ] = γπππ‘γ^(β1 ) [((1 + sinβ‘π₯ ) + (1 β sinβ‘π₯ ) + 2β((1 + sinβ‘π₯ ) (1 β sinβ‘π₯ ) ))/((β(1 + sinβ‘γπ₯ γ ))^2 β (β(1 β sinβ‘π₯ ))^2 ) ] = γπππ‘γ^(β1 ) [(1 + sinβ‘π₯ + 1 β sinβ‘π₯ + 2β((1)^2 β sin^2β‘π₯ ))/(1 + sinβ‘π₯ β 1 + sinβ‘π₯ ) ] = γπππ‘γ^(β1 ) [(2 + 2β(1 β sin^2β‘π₯ ))/(2 sinβ‘π₯ ) ] = γπππ‘γ^(β1 ) [(2 (1 + β(1 β sin^2β‘π₯ ) ) )/(2 sinβ‘π₯ )] = γπππ‘γ^(β1 ) [(1 + β(γcππ γ^2β‘π₯ ) )/sinβ‘π₯ ] (π΄π 1βγπ ππγ^2β‘π= γπππ γ^2β‘π) = γπππ‘γ^(β1 ) [(1 + cosβ‘π₯ )/sinβ‘π₯ ] = γπππ‘γ^(β1 ) [(1 + 2 cos^2β‘γπ₯/2γ β 1 )/(2 sinβ‘γ π₯/2 γ γcos γβ‘γπ₯/2γ )] = γπππ‘γ^(β1 ) [(2 cos^2β‘γπ₯/2γ )/(2 sinβ‘γ π₯/2 γ γcos γβ‘γπ₯/2γ )] = γπππ‘γ^(β1 ) [(γcos γβ‘γπ₯/2γ )/sinβ‘γ π₯/2 γ ] = γπππ‘γ^(β1 ) [cotβ‘(π₯/2) ] = π₯/2 We know that sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ Replacing ΞΈ by π/2 sin ΞΈ = 2 πππβ‘γπ½/πγ πππβ‘γπ½/πγ and cos 2ΞΈ = 2cos2 ΞΈ β 1 Replacing ΞΈ by π/2 cos ΞΈ = 2cos2 π½/π β 1 (π΄π γπππ‘γ^(β1 ) (πππ‘β‘π )=π) Thus, π= π/π Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = π/ππ₯ (π₯/2) ππ¦/ππ₯ = 1/2 ππ₯/ππ₯ π π/π π = π/π Thus, π= π/π Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = π/ππ₯ (π₯/2) ππ¦/ππ₯ = 1/2 ππ₯/ππ₯ π π/π π = π/π Thus, π= π/π Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = π/ππ₯ (π₯/2) ππ¦/ππ₯ = 1/2 ππ₯/ππ₯ π π/π π = π/π Misc 6 (Method 2) Differentiate w.r.t. x the function, γπππ‘γ^(β1 ) [(β(1 + sinβ‘π₯ ) + β(1 β sinβ‘π₯ ))/(β(1 + sinβ‘γπ₯ γ ) β β(1 β sinβ‘π₯ )) ] , 0<π₯< π/2 Let π¦= γπππ‘γ^(β1 ) [(β(1 + sinβ‘π₯ ) + β(1 β sinβ‘π₯ ))/(β(1 +γ sinγβ‘γπ₯ γ ) β β(1 β sinβ‘π₯ )) ] As γπ ππγ^2 π+γπππ γ^2β‘π=1 Replacing π by π₯/2 π ππ2 π₯/2 + γπππ γ^2 π₯/2 = 1 & π ππβ‘2π=2 π ππβ‘γπ πππ β‘π γ Replacing π by π₯/2 π ππβ‘π₯ = 2 π ππβ‘π₯/2 πππ β‘π₯/2 1 + sin x = (γπ ππγ^2 π₯/2 + γπππ γ^2 π₯/2) + 2 sin π₯/2 cos π₯/2 1 + sin x = (πππ π₯/2 +sinβ‘γπ₯/2γ )^2 β("1 + sin x" ) = (πππ π/π +πππβ‘γπ/πγ ) 1 β sin x = (γπ ππγ^2 π₯/2 + γπππ γ^2 π₯/2) β 2 sin π₯/2 cos π₯/2 1 β sin x = (πππ π₯/2 βsinβ‘γπ₯/2γ )^2 β("1 β sin x" ) = (πππ π₯/2 βsinβ‘γπ₯/2γ ) Now, y = γπππ γ^(β1) |(β(1 + sinβ‘π₯ ) + β(1 β sinβ‘π₯ ))/(β(1 + sinβ‘π₯ ) β β(1 β sinβ‘π₯ ))| Substitute value of β(1+π ππβ‘π₯ ) & β(1βπ ππβ‘π₯ ) from (1) & (2). y = cotβ1 [((γcos γβ‘γπ₯/2γ + γsin γβ‘γπ₯/2γ ) + (γcos γβ‘γπ₯/2γ β γsin γβ‘γπ₯/2γ ))/((γcos γβ‘γπ₯/2γ + γsin γβ‘γπ₯/2γ ) + (γcos γβ‘γπ₯/2γ β γsin γβ‘γπ₯/2γ ) )] y = cotβ1 [((γcos γβ‘γπ₯/2γ + γsin γβ‘γπ₯/2γ ) + (γcos γβ‘γπ₯/2γ β γsin γβ‘γπ₯/2γ ))/((γcos γβ‘γπ₯/2γ + γsin γβ‘γπ₯/2γ ) + (γcos γβ‘γπ₯/2γ β γsin γβ‘γπ₯/2γ ) )] π¦= γπππ‘γ^(β1 ) [(πππ β‘γ π₯/2γ + γπ ππ γβ‘γπ₯/2γ + πππ β‘γ π₯/2γ β γπ ππ γβ‘γπ₯/2γ)/(πππ β‘γ π₯/2γ + π ππβ‘γ π₯/2γ β γπππ γβ‘γπ₯/2γ β π ππβ‘γ π₯/2γ ) ] π¦= γπππ‘γ^(β1 ) [(2 γπππ γβ‘γπ₯/2γ )/(2 π ππβ‘γ π₯/2γ ) ] π¦= γπππ‘γ^(β1 ) [πππ‘ (π₯ )/2 ] π¦= π₯/2 (π΄π γπππ‘γ^(β1 ) (πππ‘β‘π )=π) Thus, π= π/π Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = π/ππ₯ (π₯/2) ππ¦/ππ₯ = 1/2 ππ₯/ππ₯ π π/π π = π/π
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