Example 40 (iii) - Chapter 5 Class 12 Continuity and Differentiability
Last updated at April 16, 2024 by Teachoo
Examples
Example 1
Example 2
Example 3
Example 4 Important
Example 5
Example 6
Example 7
Example 8
Example 9
Example 10
Example 11 Important
Example 12
Example 13 Important
Example 14
Example 15 Important
Example 16
Example 17 Important
Example 18
Example 19
Example 20 Important
Example 21
Example 22
Example 23
Example 24 Important
Example 25
Example 26 (i)
Example 26 (ii) Important
Example 26 (iii) Important
Example 26 (iv)
Example 27 Important
Example 28
Example 29 Important
Example 30 Important
Example 31
Example 32
Example 33 Important
Example 34 Important
Example 35
Example 36 Important
Example 37
Example 38 Important
Example 39 (i)
Example 39 (ii) Important
Example 40 (i)
Example 40 (ii) Important
Example 40 (iii) Important You are here
Example 41
Example 42 Important
Example 43
Question 1 Deleted for CBSE Board 2024 Exams
Question 2 Important Deleted for CBSE Board 2024 Exams
Question 3 Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Transcript
Example 40 Differentiate the following w.r.t. x. (iii) sin^(β1) ((2^( π₯+1) )/( 1 +γ 4 γ^π₯ )) Let π(π₯) = sin^(β1) ((2^( π₯+1) )/( 1 +γ 4 γ^π₯ )) π(π₯) = sin^(β1) ((2^( π₯). 2)/( 1 + (2^π₯ )^2 )) Let π^π = tan ΞΈ π(π₯) = sin^(β1) ((tanβ‘γπ γ. 2)/( 1 + tan^2β‘π )) = sin^(β1) ((2 tanβ‘γπ γ )/( 1 +γ tan^2γβ‘π )) = sin^(β1) (sin 2π) = 2π (sinβ‘2π "= " (2 tanβ‘π)/(1 +γ tan^2γβ‘π )) (As γπ ππγ^(β1)β‘γ(π ππβ‘π)γ =π) Since 2^π₯= tanβ‘π tan^(β1) (2^π₯ )=π β΄ π(π) = π (γπππγ^(βπ) (π^π )) Differentiating π€.π.π‘.π₯ πβ(π₯) = 2 (π (tan^(β1) 2^π₯ )" " )/ππ₯ πβ(π₯) = 2 . 1/(1 + (2^π₯ )^2 ) . (π (π^π )" " )/π π πβ(π₯) = (2 )/(1 + (2^π₯ )^2 ) . π^π . πππβ‘π (π΄π π/ππ₯(γπ‘ππγ^(β1))=1/(1+π₯^2 )) (π΄π π/ππ₯ (π^π₯ )=π^π₯. πππβ‘π₯ ) πβ(π₯) = (γ2. 2γ^π₯.γ logγβ‘2)/(1 + (2^π₯ )^2 ) πβ(π₯) = (2^(π₯ + 1).γ logγβ‘2)/(1 + (2^π₯ )^2 ) πβ(π₯) = (2^(π₯ + 1).γ logγβ‘2)/(1 + (2^2 )^π₯ ) πβ(π) = (π^(π + π).γ πππγβ‘π)/(π + π^π )
