Derivative, chain rule

Petrus

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

$\fracr} \sqrt{r^2+1}$

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

Last edited:

chisigma

Well-known member
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

Well-known member
MHB Math Helper
Re: Derivate, chain rule

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

$\fracr} \sqrt{r^2+1}$

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:

\displaystyle \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*}

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

Petrus

Well-known member
Re: Derivate, chain rule

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

\displaystyle \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*}

Then to evaluate \displaystyle \displaystyle \begin{align*} \frac{d}{dr}\left[ \left( r^2 + 1 \right)^{-\frac{1}{2}} \right] \end{align*} 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

Staff member
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

Well-known member
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

Well-known member
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

Staff member
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:

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

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

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

Petrus

Well-known member
Re: Derivate, chain rule

No, we have:

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

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

$$\displaystyle \frac{d}{dr}\left(\frac{r}{\sqrt{r^2+1}} \right)=(r^2+1)^{-\frac{3}{2}}\left((r^2+1)-r^2 \right)=(r^2+1)^{-\frac{3}{2}}$$
Thanks Mark!
I need to practice more on this!