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!

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