242.7x.25 Find the derivative of y with respect to x

[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} \\
y'&=\frac{8}{x}+?
\end{align*}

the first term was easy but the second

 

[MarkFL]

The second term is a product, so we'll need to use the Product Rule to differentiate it. The first factor could be written as:

\(\displaystyle u=\left(1-x^2\right)^{\frac{1}{2}}\)

So, to differentiate that, we'll need the Power Rule and the Chain Rule.

The second factor is:

\(\displaystyle v=\arccos(x)\implies x=\cos(v)\)

Can you use implicit differentiation on the implication to derive the needed formula?

 

[karush]

so..

$\displaystyle u'=\dfrac{x}{\sqrt{x^2+1}}$

 

[MarkFL]

so..

$\displaystyle u'=\dfrac{x}{\sqrt{x^2+1}}$

How did you obtain that result?

 

[karush]

sign error ?
$\displaystyle u=\left(1-x^2\right)^{\frac{1}{2}}\\$
$\frac{1}{2}(1-x^2)^{-1/2}(-2x)\\$
$\displaystyle u'=-\frac{2x}{2\sqrt{1-x^2}}
= -\dfrac{x}{\sqrt{1-x^2}}$

 

[MarkFL]

sign error ?
$\displaystyle u=\left(1-x^2\right)^{\frac{1}{2}}\\$
$\frac{1}{2}(1-x^2)^{-1/2}(-2x)\\$
$\displaystyle u'=-\frac{2x}{2\sqrt{1-x^2}}
= -\dfrac{x}{\sqrt{1-x^2}}$

Yes, that's much better. (Yes)

Okay, now let's look at:

\(\displaystyle x=\cos(v)\)

What do you get when you implicitly differentiate w.r.t $x$?

 

[karush]

Yes, that's much better. (Yes)

Okay, now let's look at:

\(\displaystyle x=\cos(v)\)

What do you get when you implicitly differentiate w.r.t $x$?

do you mean

where $\displaystyle u=\left(1-x^2\right)^{\frac{1}{2}}$ and $u'=-\dfrac{x}{\sqrt{1-x^2}}$
and $x=\cos(v)$ and $x'=-\sin\left(v\right)$

$$ux'+u'x$$

 

[MarkFL]

do you mean

where $\displaystyle u=\left(1-x^2\right)^{\frac{1}{2}}$ and $u'=-\dfrac{x}{\sqrt{1-x^2}}$
and $x=\cos(v)$ and $x'=-\sin\left(v\right)$

$$ux'+u'x$$

You need to treat $v$ as a function of $x$, and therefore use the chain rule...

 

[MarkFL]

Let's look at:

\(\displaystyle x=\cos(v)\)

Implicitly differentiating w.r.t $x$, we obtain:

\(\displaystyle 1=-\sin(v)\d{v}{x}\implies \d{v}{x}=-\frac{1}{\sin(v)}\)

Now, let's go back to:

\(\displaystyle v=\arccos(x)\)

We now have:

\(\displaystyle \d{v}{x}=-\frac{1}{\sin(\arccos(x))}=-\frac{1}{\sqrt{1-x^2}}\)

Okay, we are now ready to write:

\(\displaystyle \frac{d}{dx}(uv)=\d{u}{x}v+u\d{v}{x}=?\)

 

[karush]

kinda maybe??

\(\displaystyle \frac{d}{dx}(uv)=\d{u}{x}v+u\d{v}{x}=?\)

\(\displaystyle \frac{d}{dx}(uv)=\left[-\dfrac{x}{\sqrt{1-x^2}}\right] \left[\arccos(x)\right]
+\left[\left(1-x^2\right)^{\frac{1}{2}}\right]\left[ -\frac{1}{\sqrt{1-x^2}} \right]\)

 

[MarkFL]

kinda maybe??

\(\displaystyle \frac{d}{dx}(uv)=\d{u}{x}v+u\d{v}{x}=?\)

\(\displaystyle \frac{d}{dx}(uv)=\left[-\dfrac{x}{\sqrt{1-x^2}}\right] \left[\arccos(x)\right]
+\left[\left(1-x^2\right)^{\frac{1}{2}}\right]\left[ -\frac{1}{\sqrt{1-x^2}} \right]\)

Yes...can you simplify? :D

 

[karush]

is:
$\displaystyle\arccos(x) =\frac{1}{\sqrt{1-{x}^{2}}}$

 

[MarkFL]

is:
$\displaystyle\arccos(x) =\frac{1}{\sqrt{1-{x}^{2}}}$

No, but there is an algebraic simplification that can be made in the expression you posted for the derivative of the product $uv$. :D

 

[karush]

\(\displaystyle \frac{d}{dx}(uv)=\d{u}{x}v+u\d{v}{x}=?\)

\(\displaystyle \frac{d}{dx}(uv)=\left[-\dfrac{x}{\sqrt{1-x^2}}\right] \left[\arccos(x)\right]
+\left[\left(1-x^2\right)^{\frac{1}{2}}\right]\left[ -\frac{1}{\sqrt{1-x^2}} \right]\)

$y'=\dfrac{8}{x}-\dfrac{x\arccos\left(x\right)}{\sqrt{1-x^2}}-1$

not sure how this could be simplified??

 

[MarkFL]

\(\displaystyle \frac{d}{dx}(uv)=\d{u}{x}v+u\d{v}{x}=?\)

\(\displaystyle \frac{d}{dx}(uv)=\left[-\dfrac{x}{\sqrt{1-x^2}}\right] \left[\arccos(x)\right]
+\left[\left(1-x^2\right)^{\frac{1}{2}}\right]\left[ -\frac{1}{\sqrt{1-x^2}} \right]\)

$y'=\dfrac{8}{x}-\dfrac{x\arccos\left(x\right)}{\sqrt{1-x^2}}-1$

not sure how this could be simplified??

You did the simplification I was talking about...great job! (Yes)

 

[Greg]

Interesting problem!

$$\cos(u)=x$$

$$\frac{du}{dx}\sin(u)=-1$$

$$\frac{du}{dx}=-\frac{1}{\sqrt{1-x^2}}$$

(We use the positive root as $u\in[0,\pi]$)

$$y=\sqrt{1-x^2}\arccos(x)=u\sin(u)$$

$$y'=u\cos(u)\frac{du}{dx}+\sin(u)\frac{du}{dx}=\frac{du}{dx}\left(u\cos(u)+\sin(u)\right)=-\frac{x\arccos(x)}{\sqrt{1-x^2}}-1$$

 

[karush]

much mahalo everyone

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