# Thread: Cross product

1. ## Cross product derivative

Find 1st and 2nd derivative.

$\displaystyle f(t) = (\bold a \times t \bold b) \times (\bold a + t^2 \bold b)$

$\displaystyle (\bold a \times t\bold b)' = (\bold a \times \bold b)$
$\displaystyle (\bold a + t^2 \bold b)' = 2t \bold b$

so...

$\displaystyle f'(t) = (\bold a \times t\bold b) \times (\bold a + t^2\bold b)' + (\bold a \times t\bold b)' \times (\bold a + t^2\bold b)$
$\displaystyle f'(t) = (\bold a \times t\bold b) \times (2t\bold b) + (\bold a \times \bold b) \times (\bold a + t^2\bold b)$

Here's where I'm having trouble.

$\displaystyle f''(t) = [(\bold a \times t\bold b) \times (2t\bold b)' + (\bold a \times t\bold b)' \times (\bold a + t^2\bold b) ] + [(\bold a \times \bold b) \times (\bold a + t^2\bold b)' + (\bold a \times \bold b)' \times (\bold a + t^2\bold b)] \;?$

Or should I use distributive property on that 2nd part?

$\displaystyle f''(t) = [(\bold a \times t\bold b) \times (2t\bold b)' + (\bold a \times \bold b)' \times (\bold a + t^2\bold b) ] + [(\bold a \times \bold b) \times (\bold a) + (\bold a \times \bold b) \times (t^2\bold b)]$

2. ## Re: Cross product

We may write

$\displaystyle f(t) = (\bold a \times t \bold b) \times (\bold a + t^2 \bold b)=$

$\displaystyle =t \; (\bold a \times \bold b) \times \bold a+t^3\; (\bold a \times \bold b) \times \bold b.$