Finding derivative of a function by chain rule

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

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Misc 6 (Method 1) Differentiate w.r.t. x the function, γπππ‘γ^(β1 ) [(β(1 + sinβ‘π₯ ) + β(1 β sinβ‘π₯ ))/(β(1 + sinβ‘γπ₯ γ ) β β(1 βsinβ‘π₯ )) ] , 0<π₯< π/2 Let π¦= γπππ‘γ^(β1 ) [(β(1 + sinβ‘π₯ ) +β(1 β sinβ‘π₯ ))/(β(1 + sinβ‘γπ₯ γ )ββ(1 β sinβ‘π₯ )) ] Rationalizing the sum π¦= γπππ‘γ^(β1 ) [((β(1 + sinβ‘π₯ ) + β(1 β sinβ‘π₯ )))/((β(1 + sinβ‘γπ₯ γ )β β(1 β sinβ‘π₯ )) ) Γ((β(1 + sinβ‘π₯ ) + β(1 β sinβ‘π₯ )))/((β(1+sinβ‘γπ₯ γ )+ β(1 β sinβ‘π₯ )) )] π¦= γπππ‘γ^(β1 ) [(β(1 + sinβ‘π₯ ) + β(1 β sinβ‘π₯ ))^2/((β(1 + sinβ‘γπ₯ γ )β β(1 βγ sinγβ‘π₯ )) (β(1 + sinβ‘γπ₯ γ )+ β(1 βγ sinγβ‘π₯ )) ) ] = γπππ‘γ^(β1 ) [((β(1 + sinβ‘π₯ ) )^2 + (β(1 β sinβ‘π₯ ) )^2+ 2(β(1 + sinβ‘γπ₯ γ ))(β(1 βγ sinγβ‘π₯ )))/((β(1 + sinβ‘γπ₯ γ )β β(1 β γ sinγβ‘π₯ )) (β(1 + sinβ‘γπ₯ γ )+ β(1 β sinβ‘π₯ )) ) ] = γπππ‘γ^(β1 ) [((1 + sinβ‘π₯ ) + (1 β sinβ‘π₯ ) + 2β((1 + sinβ‘π₯ ) (1 β sinβ‘π₯ ) ))/((β(1 + sinβ‘γπ₯ γ ))^2 β (β(1 β sinβ‘π₯ ))^2 ) ] = γπππ‘γ^(β1 ) [(1 + sinβ‘π₯ + 1 β sinβ‘π₯ + 2β((1)^2 β sin^2β‘π₯ ))/(1 + sinβ‘π₯ β 1 + sinβ‘π₯ ) ] = γπππ‘γ^(β1 ) [(2 + 2β(1 β sin^2β‘π₯ ))/(2 sinβ‘π₯ ) ] = γπππ‘γ^(β1 ) [(2 (1 + β(π β γπππγ^πβ‘π ) ) )/(2 sinβ‘π₯ )] = γπππ‘γ^(β1 ) [(1 + β(γπππγ^πβ‘π ) )/sinβ‘π₯ ] = γπππ‘γ^(β1 ) [(1 + πππβ‘π )/πππβ‘π ] = γπππ‘γ^(β1 ) [(1 + π γπππγ^πβ‘γπ/πγ β π )/(π πππβ‘γ π/π γ γππ¨π¬ γβ‘γπ/πγ )] = γπππ‘γ^(β1 ) [(2 cos^2β‘γπ₯/2γ )/(2 sinβ‘γ π₯/2 γ γcos γβ‘γπ₯/2γ )] = γπππ‘γ^(β1 ) [(γcos γβ‘γπ₯/2γ )/sinβ‘γ π₯/2 γ ] = γπππ‘γ^(β1 ) [cotβ‘(π₯/2) ] = π₯/2 We know that sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ Replacing ΞΈ by π/2 sin ΞΈ = 2 πππβ‘γπ½/πγ πππβ‘γπ½/πγ and cos 2ΞΈ = 2cos2 ΞΈ β 1 Replacing ΞΈ by π/2 cos ΞΈ = 2cos2 π½/π β 1 Thus, π= π/π Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = π/ππ₯ (π₯/2) ππ¦/ππ₯ = 1/2 ππ₯/ππ₯ ππ/ππ = π/π Misc 6 (Method 2) Differentiate w.r.t. x the function, γπππ‘γ^(β1 ) [(β(1 + sinβ‘π₯ ) + β(1 β sinβ‘π₯ ))/(β(1 + sinβ‘γπ₯ γ ) β β(1 β sinβ‘π₯ )) ] , 0<π₯< π/2 Let π¦= γπππ‘γ^(β1 ) [(β(1 + sinβ‘π₯ ) + β(1 β sinβ‘π₯ ))/(β(1 +γ sinγβ‘γπ₯ γ ) β β(1 β sinβ‘π₯ )) ] Finding β(π + πππβ‘π ) and β(π β πππβ‘π ) γπππγ^π π½+γπππγ^πβ‘π½=1 Replacing π by π₯/2 π ππ2 π₯/2 + γπππ γ^2 π₯/2 = 1 πππβ‘ππ½=2 π ππβ‘γπ πππ β‘π γ Replacing π by π₯/2 πππβ‘π = 2 π ππβ‘π₯/2 πππ β‘π₯/2 β("1 + sin x" ) = β((γπ ππγ^2 π₯/2 + γπππ γ^2 π₯/2)" + 2 sin " π₯/2 " cos " π₯/2) = β((πππ  π₯/2 +sinβ‘γπ₯/2γ )^2 ) = πππ π/π +πππβ‘γπ/πγ β("1 " β" sin x" ) = β((γπ ππγ^2 π₯/2 + γπππ γ^2 π₯/2)" β 2 sin " π₯/2 " cos " π₯/2) = β((πππ  π₯/2 βsinβ‘γπ₯/2γ )^2 ) = πππ π/π +πππβ‘γπ/πγ β("1 " β" sin x" ) = β((γπ ππγ^2 π₯/2 + γπππ γ^2 π₯/2)" β 2 sin " π₯/2 " cos " π₯/2) = β((πππ  π₯/2 βsinβ‘γπ₯/2γ )^2 ) = πππ π/π +πππβ‘γπ/πγ Thus, our equation becomes y = γπππ γ^(β1) |(β(1 + sinβ‘π₯ ) + β(1 β sinβ‘π₯ ))/(β(1 + sinβ‘π₯ ) β β(1 β sinβ‘π₯ ))| Substituting value of β(1+π ππβ‘π₯ ) & β(1βπ ππβ‘π₯ ) from (1) & (2). y = cotβ1 [((γcos γβ‘γπ₯/2γ + γsin γβ‘γπ₯/2γ ) + (γcos γβ‘γπ₯/2γ β γsin γβ‘γπ₯/2γ ))/((γcos γβ‘γπ₯/2γ + γsin γβ‘γπ₯/2γ ) + (γcos γβ‘γπ₯/2γ β γsin γβ‘γπ₯/2γ ) )] π¦= γπππ‘γ^(β1 ) [(πππ β‘γ π₯/2γ + γπ ππ γβ‘γπ₯/2γ + πππ β‘γ π₯/2γ β γπ ππ γβ‘γπ₯/2γ)/(πππ β‘γ π₯/2γ + π ππβ‘γ π₯/2γ β γπππ  γβ‘γπ₯/2γ β π ππβ‘γ π₯/2γ ) ] π¦= γπππ‘γ^(β1 ) [(2 γπππ  γβ‘γπ₯/2γ )/(2 π ππβ‘γ π₯/2γ ) ] π¦= γπππ‘γ^(β1 ) [πππ‘ (π₯ )/2 ] π= π/π Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = π(π₯/2)/ππ₯ ππ/ππ = π/π