Ex 5.3, 14 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at March 11, 2021 by Teachoo
Last updated at March 11, 2021 by Teachoo
Transcript
Ex 5.3, 14 Find ππ¦/ππ₯ in, y = sinβ1 (2π₯ β(1βπ₯^2 )) , β 1/β2 < x < 1/β2 y = sinβ1 (2π₯ β(1βπ₯^2 )) Putting π₯ =π ππβ‘π π¦ = sinβ1 (2 sinβ‘π β(1βγπ ππγ^2 π)) π¦ = sinβ1 ( 2 sin ΞΈ β(γπππ γ^2 π)) π¦ ="sinβ1 " (γ"2 sin ΞΈ" γβ‘cosβ‘π ) π¦ = sinβ1 (sinβ‘γ2 π)γ π¦ = 2ΞΈ Putting value of ΞΈ = sinβ1 x π¦ = 2 γπ ππγ^(β1) π₯ Since x = sin ΞΈ β΄ γπ ππγ^(β1) x = ΞΈ Differentiating both sides π€.π.π‘.π₯ . (π(π¦))/ππ₯ = (π (γ2 sin^(β1)γβ‘π₯ ))/ππ₯ ππ¦/ππ₯ = 2 (πγ (π ππγ^(β1) π₯))/ππ₯ ππ¦/ππ₯ = 2 (1/β(1 βγ π₯γ^2 )) π π/π π = π/β(π β π^π ) ((sin^(β1)β‘π₯ )^β²= 1/β(1 β π₯^2 ))
Finding derivative of Inverse trigonometric functions
Example 26 Important
Example 27
Derivative of cot-1 x (cot inverse x)
Derivative of sec-1 x (Sec inverse x)
Derivative of cosec-1 x (Cosec inverse x)
Ex 5.3, 14 You are here
Ex 5.3, 9 Important
Ex 5.3, 13 Important
Ex 5.3, 12 Important
Ex 5.3, 11 Important
Ex 5.3, 10 Important
Ex 5.3, 15 Important
Misc 5 Important
Misc 4
Misc 13 Important
Finding derivative of Inverse trigonometric functions
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