Answer
$\frac{∂^{3}W}{∂u^{2}∂v}=\frac{3}{4}v (u+v^{2})^{-5/2}$
Work Step by Step
Consider the function $W=\sqrt {u+v^{2}}$
Let us start differentiating the function with respect to $v$ keeping $u$ constant.
$\frac{∂W}{∂v}=\frac{1}{2\sqrt {u+v^{2}}}\times2v $
$=v( {u+v^{2}})^{-1/2}$
Differentiate with respect to $u$ keeping $v$ constant.
$\frac{∂^{2}W}{∂u∂v}=-\frac{1}{2}v (u+v^{2})^{-3/2}$
Apply power rule.
Differentiate with respect to $u$ keeping $v$ constant.
$\frac{∂^{3}W}{∂u^{2}∂v}=-\frac{1}{2}(\frac{-3}{2})v (u+v^{2})^{-5/2}$
$=\frac{3}{4}v (u+v^{2})^{-5/2}$
Hence, $\frac{∂^{3}W}{∂u^{2}∂v}=\frac{3}{4}v (u+v^{2})^{-5/2}$
