IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - Derivatives of inverse trigonometric functions

kishore.subramanian.b

i) y= sin^-1[2x/1+x² ].

put x=tanz

then sin^-1[2x/1+x² ] =2z=2tan^-1x

hence y=2tan^-1x

dy/dx=2.1/(1+x²)

ii) y= sin^-1 [1-x² /1+x² ].

put x=tan z

hence sin^-1 [1-x² /1+x² ]=sin^-1(cos2z)=sin^-1(sin(90-2z))=90-2tan^-1x

y=90-2tan^-1x

dy/dx=-2/(1+x²)

y=sin^-1 [2x(

1-x² )]

put x=sinz

sin^-1 [2x(

1-x² )]=2z=2sin^-1x

hence y=2sin^-1x

dy/dx=2/sqrt(1+x²)

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