MHB Derivative of y w.r.t x: 242.7x.25

[karush]

$\tiny{242.7x.25}$
$\textsf{Find the derivative of y with respect to x}$
\begin{align*}\displaystyle
y&=8\ln{x}+\sqrt{1-x^2}\arccos{x} \\
&=\frac{8}{x}+?
\end{align*}

 

[MarkFL]

We could use a table of forumlas here, but let's derive the formula we need. Let:

$$y=\arccos(x)$$

$$\cos(y)=x$$

Implicitly differentiate:

$$-\sin(y)\d{y}{x}=1$$

$$\d{y}{x}=-\csc(y)=-\csc(\arccos(x))=-\frac{1}{\sqrt{1-x^2}}$$

Can you proceed with the product rule?

 

[Prove It]

We could use a table of forumlas here, but let's derive the formula we need. Let:

$$y=\arccos(x)$$

$$\cos(y)=x$$

Implicitly differentiate:

$$-\sin(y)\d{y}{x}=1$$

$$\d{y}{x}=-\csc(y)=-\csc(\arccos(x))=-\frac{1}{\sqrt{1-x^2}}$$

Can you proceed with the product rule?

As it's not entirely obvious why $\displaystyle \begin{align*} \csc{ \left[ \arccos{ \left( x \right) } \right] } = \frac{1}{\sqrt{1 - x^2}} \end{align*}$

$\displaystyle \begin{align*} \csc{ \left[ \arccos{(x)} \right] } &= \frac{1}{\sin{\left[ \arccos{ (x) } \right] }} \\ &= \frac{1}{\sqrt{ 1 - \left\{ \cos{ \left[ \arccos{(x)} \right] } \right\} ^2 }} \\ &= \frac{1}{\sqrt{1 - x^2}} \end{align*}$