## Differential coefficient of sec (γtanγ^(-1)x) w.r.t. x is

## (A) x/β(1 + x^2 )Β

## (B) x/(1 + x^2 )

## (C) x β(1+x^2 ) Β

## (D) 1/β(1 + x^2 )

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Ex 5.1, 21 - Chapter 5 Class 12 -Continuity and Differentiability

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Chapter 5 Class 12 Continuity and Differentiability (Term 1)

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This question is similar to

Ex 5.1, 21 - Chapter 5 Class 12 -Continuity and Differentiability

Hello! Teachoo has made this answer with days (even weeks!) worth of effort and love. If Teachoo has been of any help to you in your Board exam preparation, then please support us by clicking on this link to make a donation

Question 9 Differential coefficient of sec (γπ‘ππγ^(β1)x) w.r.t. x is (A) π₯/β(1 + π₯^2 ) (B) π₯/(1 + π₯^2 ) (C) x β(1+π₯^2 ) (D) 1/β(1 + π₯^2 ) Let y = sec (γπ‘ππγ^(β1) π₯) Differential coefficient sec (γπ‘ππγ^(β1)x) means π π/π π Finding π π/π π π π/π π = π(secβ‘(γπ‘ππγ^(β1) π₯))/ππ₯ = sec (γπ‘ππγ^(β1) π₯) tan (γπππγ^(βπ) π) Γ (π(γπ‘ππγ^(β1) π₯))/ππ₯ = sec (γπ‘ππγ^(β1) π₯) . π . 1/(1 + π₯^2 ) = sec(γπππγ^(βπ) π) Γ π/(π + π^π ) Now, to find π π/π π , we need to find value of sec(γπππγ^(βπ) π) Finding value of sec(γπππγ^(βπ) π) Let π½ = γπππγ^(βπ) π tanβ‘π=π₯ Now, let ABC be a triangle tanβ‘π=π΄π΅/π΅πΆ π₯=π΄π΅/π΅πΆ π΄π΅/π΅πΆ=π₯/1 β΄ AB = π & BC = 1 So, AC = β(γπ΄π΅γ^2+γπ΅πΆγ^2 ) = β(π₯^2+1^2 ) = β(π^π+π) Now, cosβ‘γπ=π΅πΆ/π΄πΆγ =1/β(π₯^2 + 1) β΄ π¬ππβ‘π½=1/cosβ‘π =1/(1/β(π₯^2+1)) =β(π^π+π) Putting π = γπππγ^(βπ) π sec (γπππγ^(βπ) π) =β(π₯^2+1) Putting value of sec (γπ‘ππγ^(β1) π₯) in (1) π π/π π = sec(γπ‘ππγ^(β1) π₯) Γ π₯/(1+ π₯^2 ) = β(π^π+π) Γ π₯/(1 + π₯^2 ) = β(π₯^2+1) Γ π₯/(β(π₯^2 + 1))^2 = π/β(π^π + π) So, the correct answer is (A)