Proof:
Let us assume [tex]y=r\cdot s[/tex]
Taking log on both sides we get
[tex]ln(y)=ln(r)+ln(s)[/tex]
[tex]\because ln(a\cdot b)=ln(a)+ln(b)[/tex]
Now differentiating on both sides and noting that [tex]d(ln(y))=\frac{1}{y}\cdot y'[/tex] we get
[tex]\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}[/tex]
