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

Sep 12, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{- x \left(x + 1\right)}}$.

Explanation:

It is known that $\frac{d}{\mathrm{dt}} {\sin}^{-} 1 t = \frac{1}{\sqrt{1 - {t}^{2}}}$.

Let $\left(2 x + 1\right) = t$

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

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

Therefore, by Chain Rule,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}} \ldots \ldots \ldots \ldots \ldots \ldots \left(\star\right)$

Now, $y = {\sin}^{-} 1 t \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}} = \frac{1}{\sqrt{1 - {t}^{2}}} \ldots \ldots \ldots \ldots \ldots . \left(1\right)$

$t = 2 x + 1 \Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}} = 2. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(2\right)$

Using $\left(1\right) , \left(2\right) \text{ in } \left(\star\right)$, &, remembering that $t = 2 x + 1$, we get,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{1}{\sqrt{1 - {t}^{2}}}\right) \left(2\right) = \frac{2}{\sqrt{1 - {\left(2 x + 1\right)}^{2}}} = \frac{2}{\sqrt{1 - 4 {x}^{2} - 4 x - 1}}$

Therefore, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{- x \left(x + 1\right)}}$.