# How do you find the derivative of y = arcsin(1 - 2sin^2x)?

Oct 10, 2017

We can rewrite as

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

Which in turn can be rewritten as

$\sin y = \cos \left(2 x\right)$

We now differentiate implicitly on the left side and using the chain rule on the right.

$\cos y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = - 2 \sin \left(2 x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 \sin \left(2 x\right)}{\cos} y$

We know that ${\sin}^{2} y + {\cos}^{2} y = 1$ and so cosy = sqrt(1 -sin^2y. From the original function, we know that $\cos y = \sqrt{1 - {\left(1 - 2 {\sin}^{2} x\right)}^{2}} = \sqrt{1 - \left(1 - 4 {\sin}^{2} x + 4 {\sin}^{4} x\right)} = \sqrt{4 {\sin}^{2} x - 4 {\sin}^{4} x}$. Accordingly,

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

We can simplify as follows:

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 \cos x}{\sqrt{{\cos}^{2} x}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 \cos x}{|} \cos x |$

Depending upon the value of $x$, $\frac{\mathrm{dy}}{\mathrm{dx}}$ will either be positive or negative.

$\cos x$ will negative whenever pi/2 ≤ x ≤ (3pi)/2.

Since the periodicity of $\cos x$ is $2 \pi$.

We can therefore define the derivative by the piecewise function

dy/dx = {(2, "when " pi/2 + npi ≤ x ≤ pi +npi, x in RR, n in ZZ), (-2, "when " npi ≤ x ≤ pi/2 +npi, x in RR, n in ZZ):}

Thanks to Douglas K for his guidance here!

Hopefully this helps!

Oct 10, 2017

y’=(-2cosx)/sqrt(1+cos^2x)

#### Explanation:

$y = {\sin}^{-} 1 \left(1 - 2 {\sin}^{2} x\right) = {\sin}^{-} 1 \left({\cos}^{2} x\right)$
Put $u = {\cos}^{2} x$
$\frac{\mathrm{du}}{\mathrm{dx}} = - 2 \cos x \cdot \sin x = - \sin 2 x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{d}{\mathrm{dx}}\right) {\sin}^{-} 1 u = \left(\frac{1}{\sqrt{1 - {u}^{2}}}\right) \frac{\mathrm{du}}{\mathrm{dx}}$
y’=(1/sqrt(1-cos^4x))*(-sin 2x)
y’=(-sin 2x)/(sqrt((1+cos^2x)(1-cos^2x)))
$= \frac{- 2 \cdot \cancel{\sin x} \cdot \cos x}{\cancel{\sin x} \cdot \sqrt{1 + {\cos}^{2} x}}$
y’=(-2cos x)/sqrt(1+cos^2x)