# Thread: Missing steps in book

1. ## Missing steps in book

Im trying to follow another example in this book, and im not sure how the following is done

$\displaystyle \frac{d^2x}{dr^2} + \frac{1}{r} * \frac{dx}{dr}- \frac{x}{r^2} = \frac{Q}{D}$

apprantley can be rearanged as follows:

$\displaystyle \frac{d}{dr} [ \frac{1}{r} * \frac{d}{dr} (xr) ] = \frac{Q}{D}$

Calypso

2. Originally Posted by calypso
Im trying to follow another example in this book, and im not sure how the following is done

$\displaystyle \frac{d^2x}{dr^2} + \frac{1}{r} * \frac{dx}{dr}- \frac{x}{r^2} = \frac{Q}{D}$

apprantley can be rearanged as follows:

$\displaystyle \frac{d}{dr} [ \frac{1}{r} * \frac{d}{dr} (xr) ] = \frac{Q}{D}$

Calypso
When we have proofs or something that simplifies and we don't know how it's done, the best thing to do is work backwards. So let us do that.

$\displaystyle \frac{d}{dr} [ \frac{1}{r} * \frac{d}{dr} (xr) ] = \frac{Q}{D}$

$\displaystyle \frac{d}{dr} [ \frac{1}{r} * (x + \frac{dx}{dr} r ) ] = \frac{Q}{D}$

$\displaystyle \frac{d}{dr} [ \frac{1}{r}x + \frac{1}{r} \frac{dx}{dr} r ] = \frac{Q}{D}$

$\displaystyle \frac{d}{dr} [ \frac{1}{r}x + \frac{dx}{dr} ] = \frac{Q}{D}$

$\displaystyle \frac{d}{dr} \frac{1}{r}x + \frac{d}{dr} \frac{dx}{dr} = \frac{Q}{D}$

$\displaystyle \frac{dx}{dr} \frac{1}{r} - \frac{1}{r^2}x + \frac{d}{dr} \frac{dx}{dr} = \frac{Q}{D}$

$\displaystyle \frac{dx}{dr} \frac{1}{r} - \frac{1}{r^2}x + \frac{d^2x}{d^2r} = \frac{Q}{D}$

3. Originally Posted by calypso
Im trying to follow another example in this book, and im not sure how the following is done

$\displaystyle \frac{d^2x}{dr^2} + \frac{1}{r} * \frac{dx}{dr}- \frac{x}{r^2} = \frac{Q}{D}$

apprantley can be rearanged as follows:

$\displaystyle \frac{d}{dr} [ \frac{1}{r} * \frac{d}{dr} (xr) ] = \frac{Q}{D}$

Calypso
Note the last two terms on the LHS

$\displaystyle \frac{1}{r} \frac{dx}{dr}- \frac{x}{r^2} = \frac{d}{dr} \left( \frac{x}{r} \right)$

so all three terms become

$\displaystyle \frac{d^2x}{dr^2} + \frac{d}{dr} \left( \frac{x}{r} \right) = \frac{d}{dr} \left(\frac{dx}{dr} + \frac{x}{r} \right) = \frac{d}{dr} \left(\frac{r\frac{dx}{dr} + x}{r} \right)$