What is the derivative of y=arcsin(2x+1)?

Jun 7, 2017

The derivative is $= \frac{1}{\sqrt{- x \left(x + 1\right)}}$

Explanation:

$y = \arcsin \left(2 x + 1\right)$

So,

$\sin y = 2 x + 1$

Differentiating wrt $x$

$\cos y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{\cos} y$

${\sin}^{2} y + {\cos}^{2} y = 1$

${\cos}^{2} y = 1 - {\sin}^{2} y$

$= 1 - {\left(2 x + 1\right)}^{2}$

$= 1 - 4 {x}^{2} - 4 x - 1$

$= - 4 \left({x}^{2} + x\right)$

$\cos y = \sqrt{- 4 \left({x}^{2} + x\right)}$

Therefore,

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

$= \frac{1}{\sqrt{- x \left(x + 1\right)}}$

Jun 7, 2017

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

Explanation:

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \frac{d}{\mathrm{dx}} \left({\sin}^{-} 1 x\right) = \frac{1}{\sqrt{1 - {x}^{2}}}$

$\text{differentiate using the "color(blue)"chain rule}$

• d/dx(sin^-1(f(x)))=1/sqrt(1-(f(x))^2)xxf'(x)

$y = {\sin}^{-} 1 \left(2 x + 1\right)$

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

$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{2}{\sqrt{1 - {\left(2 x + 1\right)}^{2}}}$