Verified answer

Method - 1 : Chain Rule

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Let : y = tanֿ¹ (u), where : u = √(8x²-1).

Then : dy/du = 1/ (1+u²) = 1 / [ 1 + (8x²-1) ] = 1/ (8x²) ...... (1)

Also : du/dx = 1/[2√(8x²-1)]· d/dx(8x²-1)

.. . . . . . . . . = 1/[2√(8x²-1)]· 8(2x)

.. . . . . . . . . = 8x / √(8x²-1) ............................................... (2)

From (1), (2) and by Chain Rule,

dy/dx = ( dy/du ) • ( du/dx )

. . . . .= [ 1 / (8x²) ] • [ 8x / √(8x²-1) ]

. . . . .= 1 / [ x·√(8x²-1) ] .............................. Ans.

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Method - 2 : Substitution

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Let : y = tanֿ¹ [ √(8x² - 1) ],

i.e. : y = tanֿ¹ { √[ (2√2.x)² - 1] } ............................... (1)

Put : 2√2.x = sec θ so that θ = secֿ¹ (2√2.x).

Then : √[ (2√2.x)² - 1] = √(sec² θ - 1) = √(tan² θ) = tan θ.

Hence, from (1),

y = tanֿ¹ ( tan θ ) = θ = secֿ¹ (2√2.x)

Hence, by Chain Rule,

... dy/dx

= { 1 / [ (2√2.x) √((2√2.x)² - 1) ] } • d/dx ( 2√2. x )

= { 1 / [ (2√2.x) √(8x²-1) ] } • 2√2.(1)

= 1 / [ x·√(8x²-1) ] .......................................................... Ans.

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instead of arctanx use tan^(-1)x which is NOT 1/tanx !!

dy/dx=1/(1+(8x^2-1)^1/2)*16x/2(8x^2-1)^(1/2)=1/x(8x^2-1)^(1/2)