I can't seem to undersatnd how did we compute $\frac{}{}$ from the function $\mathbf T=\frac{v }{1+uvw}$ $$ \mathbf T = \frac{v}{1+uvw}$$
Solution: 
$$d\mathbf T =
\frac{\partial\mathbf T} {\partial u}\,du +\frac{\partial\mathbf T}{\partial v} \, dv +\frac{\partial\mathbf T}{\partial w} \, dw $$
$$d\mathbf T =
\frac{−v^2w }{(1+uvw)^2}\,du +
\frac{1}{(1+uvw)^2} \, dv +\frac{−uv^2}{(1+uvw)^2} \, dw$$
I get this  instead of the above solution in computing:
$$\frac{\partial\mathbf T}{\partial v} = \frac{-vuw}{(1+uvw)^2}+\frac{1}{(1+uvw)}$$

Quotient rule:
\begin{align}
& \frac\partial {\partial v} \, \frac v {1+uvw} = \frac{(1+uvw) \dfrac\partial{\partial v} v - v \dfrac\partial {\partial v} (1+uvw)}{(1+uvw)^2} \\[12pt]
= {} & \frac{(1+uvw) - v(uw)}{(1+uvw)^2} = \frac 1 {(1+uvw)^2}.
\end{align}
