# How do you solve for the derivative of y=s(1-s^2)^(1/2)+arccos(s)?

Aug 22, 2015

y^' = -(2s^2)/(sqrt(1-s^2)

#### Explanation:

You can actually have two approaches to this problem, depending on whether or not you know what the value of $\frac{d}{\mathrm{dx}} \left(\arccos x\right)$ is.

To make the calculations more interesting (I don't recommend doing things the hard way), I'll assume that you don't know what the derivative of $\arccos x$ is. This means that you will have to use implicit differentiation to find the derivative of $y$.

So, start from the original function

$y = s {\left(1 - {s}^{2}\right)}^{\frac{1}{2}} + \arccos \left(s\right)$

Rearrange to isolate $\arccos \left(s\right)$ on one side of the equation

$\arccos \left(s\right) = y - s {\left(1 - {s}^{2}\right)}^{\frac{1}{2}}$

This is equivalent to

$s = \cos \left(y - s {\left(1 - {s}^{2}\right)}^{\frac{1}{2}}\right) \text{ } \textcolor{\mathmr{and} a n \ge}{\left(1\right)}$

Now differentiate both sides with respect to $s$. Use the chain rule for $\cos u$, with $u = y - s {\left(10 - {s}^{2}\right)}^{\frac{1}{2}}$

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

$1 = - \sin u \cdot \frac{d}{\mathrm{ds}} \left(y - s {\left(1 - {s}^{2}\right)}^{\frac{1}{2}}\right) \text{ } \textcolor{\mathmr{and} a n \ge}{\left(2\right)}$

Now focus on finding

$\frac{d}{\mathrm{ds}} \left(y - s {\left(1 - {s}^{2}\right)}^{\frac{1}{2}}\right) = \frac{d}{\mathrm{ds}} \left(y\right) - \frac{d}{\mathrm{ds}} \left(s {\left(1 - {s}^{2}\right)}^{\frac{1}{2}}\right) \text{ } \textcolor{\mathmr{and} a n \ge}{\left(3\right)}$

Use the product rule and the chain rule to calculate

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

$\frac{d}{\mathrm{ds}} \left(s {\left(1 - {s}^{2}\right)}^{\frac{1}{2}}\right) = 1 \cdot {\left(1 - {s}^{2}\right)}^{\frac{1}{2}} + s \cdot \left[\frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} \cdot {\left(1 - {s}^{2}\right)}^{- \frac{1}{2}} \cdot \left(- \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} s\right)\right]$

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

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

Plug this result into $\textcolor{\mathmr{and} a n \ge}{\left(3\right)}$

$\frac{d}{\mathrm{ds}} \left(y - s {\left(1 - {s}^{2}\right)}^{\frac{1}{2}}\right) = 1 \cdot \frac{\mathrm{dy}}{\mathrm{dx}} - {\left(1 - {s}^{2}\right)}^{- \frac{1}{2}} \cdot \left(1 - 2 {s}^{2}\right)$

Now thake this back into $\textcolor{\mathmr{and} a n \ge}{\left(2\right)}$ to get

$1 = - \sin \left(y - s {\left(1 - {s}^{2}\right)}^{\frac{1}{2}}\right) \cdot \left[\frac{\mathrm{dy}}{\mathrm{dx}} - \left(1 - 2 {s}^{2}\right) {\left(1 - {s}^{2}\right)}^{- \frac{1}{2}}\right]$

Use the fact that $\textcolor{b l u e}{{\sin}^{2} x + {\cos}^{2} x = 1}$ to write $\sin x$ as a function of $\cos x$.

${\sin}^{2} = 1 - {\cos}^{2} x \implies \sin x = \sqrt{1 - {\cos}^{2} x}$

This means that you have

$1 = - \sqrt{1 - {\cos}^{2} \left(y - s {\left(1 - {s}^{2}\right)}^{\frac{1}{2}}\right)} \cdot \left[\frac{\mathrm{dy}}{\mathrm{dx}} - \left(1 - 2 {s}^{2}\right) {\left(1 - {s}^{2}\right)}^{- \frac{1}{2}}\right]$

Use $\textcolor{\mathmr{and} a n \ge}{\left(1\right)}$ to write

$1 = - \sqrt{1 - {s}^{2}} \cdot \left[\frac{\mathrm{dy}}{\mathrm{dx}} - \left(1 - 2 {s}^{2}\right) {\left(1 - {s}^{2}\right)}^{- \frac{1}{2}}\right]$

This is equivalent to

$1 = - \sqrt{1 - {s}^{2}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{\sqrt{1 - {s}^{2}}}}} \cdot \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\sqrt{1 - {s}^{2}}}}}} \cdot \left(1 - 2 {s}^{2}\right)$

Isolate $\sqrt{1 - {s}^{2}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$ on one side of the equation to get

$\sqrt{1 - {s}^{2}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}} - 2 {s}^{2} - \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}}$

Finally, you have

$\frac{\mathrm{dy}}{\mathrm{dx}} = \textcolor{g r e e n}{- \frac{2 {s}^{2}}{\sqrt{1 - {s}^{2}}}}$