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

Answer 1

#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))#.
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Answer 2

To differentiate ( y = \sin^{-1}(2x + 1) ), use the chain rule:

  1. Differentiate the outer function (\sin^{-1}(u)) with respect to its inner function (u), which is (2x + 1). This derivative is ( \frac{1}{\sqrt{1 - u^2}} ).
  2. Then, differentiate the inner function (2x + 1) with respect to (x), which yields (2).

So, the derivative of ( y = \sin^{-1}(2x + 1) ) is ( \frac{2}{\sqrt{1 - (2x + 1)^2}} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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