MHB Is the chain rule necessary for finding the derivative of r/sqrt(r^2+1)?

[Petrus]

Hello,
I got problem again with chain rule and would like to have advice for this problem,

$\frac{\displaystyle r} {\displaystyle \sqrt{r^2+1}}$

is it product rule I shall also use because I have rewrite it as
$r(r^2+1)^{-0.5}$

 

[chisigma]

Re: Derivate, chain rule

The product rule is very effective also when You have to evaluate an expression like...

$\displaystyle \frac{d}{d x} \frac{f(x)}{g(x)} = \frac{f^{\ '}(x)}{g(x)} - \frac{f(x)\ g^{\ '}(x)}{g^{2}(x)}$ (1) ... expecially, as in my case, memory isn't Your best quality...

Kind regards $\chi$ $\sigma$

 

[Prove It]

Re: Derivate, chain rule

Hello,
I got problem again with chain rule and would like to have advice for this problem,

$\frac{\displaystyle r} {\displaystyle \sqrt{r^2+1}}$

is it product rule I shall also use because I have rewrite it as
$r(r^2+1)^{-0.5}$

Yes, what you have suggested is fine. Using the product rule:

[math]\displaystyle \begin{align*} \frac{d}{dr} \left[ r\left( r^2 + 1 \right)^{-\frac{1}{2}} \right] &= r\,\frac{d}{dr} \left[ \left( r^2 + 1 \right)^{-\frac{1}{2}} \right] + \left( r^2 + 1 \right)^{-\frac{1}{2}}\,\frac{d}{dr} \left( r \right) \end{align*}[/math]

Then to evaluate [math]\displaystyle \begin{align*} \frac{d}{dr}\left[ \left( r^2 + 1 \right)^{-\frac{1}{2}} \right] \end{align*}[/math] you will need to use the Chain Rule.

 

[Petrus]

Re: Derivate, chain rule

Yes, what you have suggested is fine. Using the product rule:

[math]\displaystyle \begin{align*} \frac{d}{dr} \left[ r\left( r^2 + 1 \right)^{-\frac{1}{2}} \right] &= r\,\frac{d}{dr} \left[ \left( r^2 + 1 \right)^{-\frac{1}{2}} \right] + \left( r^2 + 1 \right)^{-\frac{1}{2}}\,\frac{d}{dr} \left( r \right) \end{align*}[/math]

Then to evaluate [math]\displaystyle \begin{align*} \frac{d}{dr}\left[ \left( r^2 + 1 \right)^{-\frac{1}{2}} \right] \end{align*}[/math] you will need to use the Chain Rule.

Hello with your method is this correcT?
$(r^2+1)^{-0.5}-0.5r(r^2+1)^{-3/2}2r$

 

[MarkFL]

Re: Derivate, chain rule

Hello with your method is this correcT?
$(r^2+1)^{-0.5}-0.5r(r^2+1)^{-3/2}2r$

Yes, and now you will probably want to factor to write the derivative in a simpler, more compact form.

 

[Petrus]

Re: Derivate, chain rule

The product rule is very effective also when You have to evaluate an expression like...

$\displaystyle \frac{d}{d x} \frac{f(x)}{g(x)} = \frac{f^{\ '}(x)}{g(x)} - \frac{f(x)\ g^{\ '}(x)}{g^{2}(x)}$ (1) ... expecially, as in my case, memory isn't Your best quality...

Kind regards $\chi$ $\sigma$

Hello Chisigma,
Is this correct?
$\frac{\sqrt{r^2+1}+0.5r(r^2+1)^{-3/2}2r}{r^2+1}$

 

[Petrus]

Re: Derivate, chain rule

Yes, and now you will probably want to factor to write the derivative in a simpler, more compact form.

Is this correct? I got problem to simplify
$(r^2+1)^{-0.5}(1-1^{4/3}r^2)$

 

[MarkFL]

Re: Derivate, chain rule

Is this correct? I got problem to simplify
$(r^2+1)^{-0.5}(1-1^{4/3}r^2)$

No, we have:

$$\frac{d}{dr}\left(\frac{r}{\sqrt{r^2+1}} \right)=(r^2+1)^{-\frac{1}{2}}-r\cdot\frac{1}{2}(r^2+1)^{-\frac{3}{2}}\cdot2r=(r^2+1)^{-\frac{1}{2}}-r^2(r^2+1)^{-\frac{3}{2}}$$

Factoring out $$(r^2+1)^{-\frac{3}{2}}$$, we find:

$$\frac{d}{dr}\left(\frac{r}{\sqrt{r^2+1}} \right)=(r^2+1)^{-\frac{3}{2}}\left((r^2+1)-r^2 \right)=(r^2+1)^{-\frac{3}{2}}$$

 

[Petrus]

Re: Derivate, chain rule

No, we have:

$$\frac{d}{dr}\left(\frac{r}{\sqrt{r^2+1}} \right)=(r^2+1)^{-\frac{1}{2}}-r\cdot\frac{1}{2}(r^2+1)^{-\frac{3}{2}}\cdot2r=(r^2+1)^{-\frac{1}{2}}-r^2(r^2+1)^{-\frac{3}{2}}$$

Factoring out $$(r^2+1)^{-\frac{3}{2}}$$, we find:

$$\frac{d}{dr}\left(\frac{r}{\sqrt{r^2+1}} \right)=(r^2+1)^{-\frac{3}{2}}\left((r^2+1)-r^2 \right)=(r^2+1)^{-\frac{3}{2}}$$

Thanks Mark!
I need to practice more on this!