How do you differentiate #y=sin^-1(2x+1)#?

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Answer 1 · 2016-09-12T14:12:13

#dy/dx=1/sqrt(-x(x+1))#.

It is known that #d/dtsin^-1 t=1/sqrt(1-t^2)#.

Let #(2x+1)=t#

#;. y=sin^-1 t, t=2x+1#.

Thus, #y# is a function of #t#, and, #t# of #x#.

Therefore, by Chain Rule,

#dy/dx=dy/dt*dt/dx..................(star)#

Now, #y=sin^-1t rArr dy/dt=1/sqrt(1-t^2)................(1)#

# t=2x+1 rArr dt/dx=2...............................................(2)#

Using #(1), (2)" in "(star)#, &, remembering that #t=2x+1#, we get,

#dy/dx=(1/sqrt(1-t^2))(2)=2/sqrt(1-(2x+1)^2)=2/sqrt(1-4x^2-4x-1)#

Therefore, #dy/dx=1/sqrt(-x(x+1))#.