Differentiation of Inverse trignometric function

Differentiate Arc Sin(2^x+1) / (1+4^x) with respect to x

y = sinֿ¹ [ (2^(x+1)) / ( 1 + (4^x)) ]

y = sinֿ¹ [ 2( 2^x ) / ( 1 + (2^x)²) ] ............... (1)
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Let : (2^x) = tan Φ, i.e., Φ = tanֿ¹ (2^x).

Then :

2(2^x) / ( 1 + (2^x)² ) = ( 2 tan Φ ) / ( 1 + tan² Φ ) = sin ( 2Φ ) .............. (2)
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From (1) and (2),

y = sinֿ¹ [ sin ( 2Φ ) ] = 2( Φ ) = 2· tanֿ¹ (2^x) ............................. (3)
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Diff... w.r.t. x, by Chain Rule,

dy/dx = 2· d/dx( tanֿ¹ (2^x))

. . . . .= 2· [ 1 / ( 1 + (2^x)²) ]· d/dx( 2^x )

. . . . .= 2· [ 1 / ( 1 + 4^x ) ]· (2^x)· ( ln 2 )

. . . . .= [ 2^(x+1) · ln 2 ] / ( 1 + 4^x ) ............................. Ans.
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The derivative only exists where the function is defined, and since 2^x + 1 > 1 for all real x, the arcsine is undefined everywhere. That expression doesn't describe a real function of a real variable. There's no function so there's no derivative.

i believe you asking arc sin((2^x+1) / (1+4^x))

remember diff arc sin x = 1/√(1-x²)

you can let one variable i.e u =((2^x+1) / (1+4^x))
therefore you got y= arc sin u

using chain rules dy/du *du/dx

good luck dude!