I can't seem to undersatnd how did we compute $\frac{}{}$ from the function $\mathbf T=\frac{v }{1+uvw}$

Question

$$ \mathbf T = \frac{v}{1+uvw}$$

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:

$$\frac{\partial\mathbf T}{\partial v} = \frac{-vuw}{(1+uvw)^2}+\frac{1}{(1+uvw)}$$

Answer 1

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}