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 $d(ln(y))=\frac{1}{y}\cdot y'$ we get

$\frac{1}{y}\cdot y'=\frac{1}{r}\cdot r'+\frac{1}{s}\cdot s'\\\\\therefore y'=\frac{y}{r}\cdot r'+\frac{y}{s}\cdot s'\\\\y'=\frac{rs}{r}\cdot r'+\frac{rs}{s}\cdot s'\\\\\therefore y'=s\cdot r'+r\cdot s'\\\\\frac{d(rs)}{dt}=s\times \frac{dr}{dt}+r\times \frac{ds}{dt}$