# How do you differentiate y=ssqrt(1-s^2)+arccoss?

Oct 30, 2017

$\frac{\mathrm{dy}}{\mathrm{ds}} = \frac{- 2 {s}^{2}}{\sqrt{1 - {s}^{2}}}$

#### Explanation:

Start by splitting into two separate bits and differentiating them one at a time (sorry for sticking in random letters all over the place - I have tried to avoid $y$, where a $y$ is misleading!).

Suppose $k = s \cdot \sqrt{1 - {s}^{2}}$ and $m = \arccos \left(s\right)$ and $u = \sqrt{1 - {s}^{2}}$

First $\frac{d}{\mathrm{ds}} \left(s \cdot \sqrt{1 - {s}^{2}}\right)$, is determined using the chain rule and product rule.

$\frac{\mathrm{dk}}{\mathrm{ds}} = s \cdot \frac{d}{\mathrm{ds}} \left(u \left(s\right)\right) + u \cdot \frac{d}{\mathrm{ds}} \left(s\right)$

Set $t = 1 - {s}^{2}$ and $u = {t}^{- \frac{1}{2}}$ then apply the chain rule $\frac{\mathrm{du}}{\mathrm{ds}} = \frac{\mathrm{du}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{ds}}$.

Hence, $\frac{\mathrm{dy}}{\mathrm{ds}} = - 2 s \cdot \frac{1}{2} \cdot {\left(1 - {s}^{2}\right)}^{- \frac{1}{2}}$
This simplifies down to (du)/(ds) = -s/(sqrt(1-s^2).

Next, to apply the product rule to find the overall derivative.

We have $s$ times that derivative, and the derivative of $s$ times the undifferentiated expression:

$\frac{d}{\mathrm{ds}} \left(s \cdot \sqrt{1 - {s}^{2}}\right) = \frac{- {s}^{2}}{\sqrt{1 - {s}^{2}}} + \sqrt{1 - {s}^{2}}$.

When you find a common denominator by algebraic manipulation, hence:

$\frac{\mathrm{dk}}{\mathrm{ds}} = - \frac{2 {s}^{2} - 1}{\sqrt{1 - {s}^{2}}}$.

Then, there is the differentiation of the $\arccos$ function.

Apply $\cos$ to both sides to undo the the effect of the inverse $\cos$. Hence $m = \arccos \left(s\right)$ becomes $s = \cos \left(m\right)$.

Using the knowledge that $\frac{d}{\mathrm{dx}} \left(\cos \left(x\right)\right) = - \sin \left(x\right)$, we can say that $\frac{\mathrm{ds}}{\mathrm{dm}} = - \sin \left(m\right)$

Next, to remove the necessity of the $\sin$ function in there, we can use the Pythagorean identity to find the derivative in terms of $s$ only.

Since ${\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right) = 1$, it follows that $\sin \left(\theta\right) = \sqrt{1 - {\cos}^{2} \left(\theta\right)}$.

Hence, (ds)/(dm) = -sqrt(1-s^2, because $s = \cos \left(m\right)$.

We know from the chain rule that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\frac{\mathrm{dx}}{\mathrm{dy}}}$, so finally $\frac{d}{\mathrm{ds}} \left(\arccos \left(s\right)\right) = - \frac{1}{\sqrt{1 - {s}^{2}}}$.

The simplification of the two together is quite delightful, as the already have the same denominator - so you can just add!

Finally, we have $\frac{\mathrm{dy}}{\mathrm{ds}} = - \left(\frac{2 {s}^{2}}{\sqrt{1 - {s}^{2}}}\right)$.

Sorry for all the confusion here, I hope it helps!