time derivative of work energy theorem

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so I stumbled upon a step in my textbook which I can't do by myself, hopefully someone can help me :).

It goes as follows

$${\frac{d \bf{p}}{dt} \cdot \bf{u}} = \frac{d}{dt}(\frac{m \bf{u}}{\sqrt{1-\frac{u^2}{c^2}}})\cdot \bf{u}$$ $$=\frac{m \bf{u}}{(1-\frac{u^2}{c^2})^{\frac{3}{2}}}\cdot \frac{d \bf{u}}{dt}$$

where $u$ is the velocity and $p$ the momentum as defined above.

My approach: I used the quotient rule on p: $\frac{f(x)}{g(x)}=\frac{g(x) f(x)^{\prime}-f(x) g(x)^{\prime}}{|g(x)|^{2}}$ but my expression is far from the result given, so I think I'm missing something. If you can help me it would be very helpful!

my expression: $$\frac{m \dot{u} \sqrt{1-\frac{u^{2}}{c^{2}}}+m u\left(\frac{u \dot{u}}{c^{2} \sqrt{1-\frac{u^{2}}{c^{2}}}}\right)}{1-\frac{u^{2}}{c^{2}}} \cdot u$$

Thanks in advance!