# Derivative of a Cross Product

1. Sep 9, 2013

### vroomba03

1. The problem statement, all variables and given/known data
Assume that you are given differentiable function f(t) and g(t). Find a formula for the
derivative of the cross product u(f(t)) x v(g(t)).

2. Relevant equations
d/dt(u(t) x v(t)) = (u'(t) x v(t) + u(t) x v'(t)

3. The attempt at a solution
So in this case I was thinking that you would just substitute f(t) and g(t) where t would be in the regular equation, so it would be U'(f(t)) x v(g(t)) + u(f(t)) x v'(g(t)), for the equation, but I have a feeling that thats not right just because it seems too simple.

2. Sep 9, 2013

### CAF123

u'(f(t)) = d u(f(t))/f(t) and similar result for v'(g(t)).
What you want is d/dt ( u(f(t)) × v(g(t)) ), so you will have to use chain rule.

3. Sep 9, 2013

### vroomba03

So then the equation would be U'(du(f(t))/f(t)) x V(g(t)) + U(g(t)) x V'(du(g(t))/g(t)) x U(f(t)) ?

4. Sep 9, 2013

### CAF123

No, start by applying the chain rule to find $$\frac{d}{dt} u(f(t))$$

5. Sep 9, 2013

### vroomba03

Oh okay, so that is u'(f(t))(f'(t)) when using chain rule. So you said u'(f(t)) = d u(f(t))/f(t), so then would I divide what i got by the chain rule and divide it by f(t) and that would be my u'(f(t))?

6. Sep 9, 2013

### CAF123

Yes.

You need to find $$\frac{d}{dt} \left(u(f(t)) \times v(g(t))\right) = \frac{d}{dt} u(f(t)) \times v(g(t)) + u(f(t)) \times \frac{d}{dt} v(g(t))$$

You correctly found $\frac{d}{dt} u(f(t))$. Now find $\frac{d}{dt} v(g(t))$ and substitute in.

What I should have wrote is u'(f(t)) $\equiv$ d u(f(t))/f(t), these two expressions denote the derivative of u with respect to f(t).

7. Sep 9, 2013

### vroomba03

So the final equation for the question would be u'(f(t))(f'(t)) x v(g(t)) + u(f(t)) x v'(f(t)(f'(t)) ?

8. Sep 9, 2013

### CAF123

Check the last term again. v is a function of g(t), not f(t).

9. Sep 9, 2013

### vroomba03

Oops silly error. u'(f(t))(f'(t)) x v(g(t)) + u(f(t)) x v'(g(t)(g'(t)) correct?

10. Sep 9, 2013

### CAF123

Correct

11. Sep 9, 2013

### vroomba03

Thank you a bunch!