Derivative of Cross Product with Differentiable Functions

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SUMMARY

The derivative of the cross product of two differentiable functions, u(f(t)) and v(g(t)), is calculated using the chain rule. The correct formula is given by the expression: u'(f(t))(f'(t)) x v(g(t)) + u(f(t)) x v'(g(t))(g'(t)). This formula incorporates the derivatives of both functions with respect to their respective variables, ensuring accurate differentiation of the cross product.

PREREQUISITES
  • Understanding of differentiable functions
  • Knowledge of the chain rule in calculus
  • Familiarity with vector cross products
  • Ability to compute derivatives of composite functions
NEXT STEPS
  • Study the application of the chain rule in multivariable calculus
  • Learn about vector calculus and its operations, including cross products
  • Explore differentiation techniques for composite functions
  • Practice problems involving derivatives of vector functions
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Students studying calculus, particularly those focusing on vector calculus, as well as educators and professionals seeking to deepen their understanding of derivatives in the context of cross products.

vroomba03
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Homework Statement


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)).


Homework Equations


d/dt(u(t) x v(t)) = (u'(t) x v(t) + u(t) x v'(t)

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 that's not right just because it seems too simple.
 
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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.
 
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)) ?
 
vroomba03 said:
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)) ?
No, start by applying the chain rule to find $$\frac{d}{dt} u(f(t))$$
 
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))?
 
vroomba03 said:
Oh okay, so that is u'(f(t))(f'(t)) when using chain rule.
Yes.

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))?
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).
 
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)) ?
 
vroomba03 said:
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)) ?

Check the last term again. v is a function of g(t), not f(t).
 
Oops silly error. u'(f(t))(f'(t)) x v(g(t)) + u(f(t)) x v'(g(t)(g'(t)) correct?
 
  • #10
vroomba03 said:
Oops silly error. u'(f(t))(f'(t)) x v(g(t)) + u(f(t)) x v'(g(t)(g'(t)) correct?
Correct :smile:
 
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  • #11
Thank you a bunch!
 

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