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

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