Find \frac{\mathrm{d} y}{\mathrm{d} x}  if y=\sin^{-1}\left [ \frac{2^{x+1}}{1+4^x} \right ]

 

 

 

 
 
 
 
 

Answers (1)

y=\sin^{-1}\left ( \frac{2^{x+1}}{1+4^x} \right )

y=\sin^{-1}\left ( \frac{2.2^x}{1+(2^2)^x} \right )    2= \tan \theta \: \: \: \:; \: \theta = \tan^{-1}(2x)

y=\sin^{-1}\left ( \frac{2.\tan \theta}{1+\tan^2 \theta} \right )

y=\sin^{-1}\left ( \sin 2 \theta \right )

y= 2 \theta= 2 \tan^{-1}(2^x)

\frac{\mathrm{d} y}{\mathrm{d} x}=2.\frac{1}{1+(2^x)^2}.2^x\log 2

\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2^{x+1}}{1+(4^x)} \log 2

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