Question

If r and s are vectors that depend on time, prove that the product rule for differentiating products applies to r.s, that is, that d/dt (r.s) = r· ds/dt + dr/dt ·s.

Answer 1

Proof:

Let us assume
`y=r\cdot s`

Taking log on both sides we get

`ln(y)=ln(r)+ln(s)`

`\because ln(a\cdot b)=ln(a)+ln(b)`

Now differentiating on both sides and noting that
`d(ln(y))=(1)/(y)\cdot y'` we get

`(1)/(y)\cdot y'=(1)/(r)\cdot r'+(1)/(s)\cdot s'\\\\\therefore y'=(y)/(r)\cdot r'+(y)/(s)\cdot s'\\\\y'=(rs)/(r)\cdot r'+(rs)/(s)\cdot s'\\\\\therefore y'=s\cdot r'+r\cdot s'\\\\(d(rs))/(dt)=s* (dr)/(dt)+r* (ds)/(dt)`