## Tripping over a basic derivation

This is a classic example of how I get tripped up.   What is wrong here?  (Goldstein derivation 1.1)

Show that for a single particle with constant mass the equation of motion implies the following differential equation for the kinetic energy: ${dT \over dt} = \bf{F} \cdot \bf{v}$ while if the mass varies with time the corresponding equation is ${d(mT) \over dt} = \bf{F} \cdot \bf{p}$

I find the first just fine

${dT \over dt } = {d \over dt} (1/2 m v^2) = {m \over 2} {d \over dt} ( v^2)$

$= {m \over 2} {d \over dt} ( \bf{v} \cdot \bf{v} ) = {m \over 2} 2 \bf{v} \cdot {d \bf{v} \over dt} = \bf{F} \cdot \bf{v}$

The second, I got trapped

${d (mT) \over dt} = {d (m {1 \over 2} m v^2 ) \over dt} = {1 \over 2 } {d (m^2 v^2) \over dt} = {1 \over 2} {d ( \bf{p} \cdot \bf{p} ) \over dt} = \bf{p} \cdot \bf{ \dot p} = \bf{F} \cdot \bf {p}$

Is the right answer, but I trapped myself with

${d(mT) \over dt} = m {dT \over dt} + T {dm \over dt} = m( \bf{F} \cdot \bf{v} ) + T {dm \over dt} = \bf{F} \cdot \bf{p} + T {dm \over dt }$

What is wrong?  I used the constant $m$ relation found initially in the non-constant derivation!