Proving vector derivatives and the BAC-CAB rule

[mohdfasieh]

Plz please please do reply(VECTORS related problem)

Hi.,
Well can u help me inthis problem.Bcoz after one day i am having a Physics test so please do reply .

Q: prove the following

1) d/du(A+B)=d/du(A) +d/du(B)

2) d/du(cA)=d/du c(A) +c d/du(A)

3) d/du(A.B)=d/du(A).B +A. d/du(B)

Q: prove the vector identity

A x (B x C)=B(A.C)-C(A.B)

 

[arildno]

Welcome to PF!
What have you done so far?

 

[thepatientmental]

Hi there,

I am happy to help with your vectors problem. Let's start by proving the given identities:

1) d/du(A+B) = d/du(A) + d/du(B)

To prove this, we will use the definition of derivative:

d/du(A+B) = lim(h->0) [ (A+B)(u+h) - (A+B)(u) ] / h

= lim(h->0) [ (A(u+h)+B(u+h)) - (A(u)+B(u)) ] / h

= lim(h->0) [ (A(u+h)-A(u)) + (B(u+h)-B(u)) ] / h

= lim(h->0) [ (A(u+h)-A(u)) / h + (B(u+h)-B(u)) / h ]

= d/du(A) + d/du(B)

2) d/du(cA) = d/du(c(A)) + c d/du(A)

Using the definition of derivative again, we have:

d/du(cA) = lim(h->0) [ c(A(u+h)) - c(A(u)) ] / h

= lim(h->0) [ c(A(u+h)-A(u)) ] / h

= c lim(h->0) [ (A(u+h)-A(u)) / h ]

= c d/du(A)

Also, d/du(c(A)) = d/du(c)A = c d/du(A) [using the product rule]

Therefore, d/du(cA) = c d/du(A) + d/du(c(A))

3) d/du(A.B) = d/du(A).B + A.d/du(B)

Using the definition of derivative once again, we have:

d/du(A.B) = lim(h->0) [ (A(u+h).B(u+h)) - (A(u).B(u)) ] / h

= lim(h->0) [ (A(u+h)-A(u)).B(u+h) + A(u).(B(u+h)-B(u)) ] / h

= lim(h->0) [ (A(u+h)-A(u)) / h . B(u+h) + A(u) . (B(u+h)-B(u)) / h ]

= d/du(A) . B + A . d/du(B)

Now, let's move on to proving