## Calculus 8th Edition

$\frac{∂^{3}V}{∂r∂s∂t}=\frac{12st^{2}}{(r+s^{2}+t^{3})^{3}}$
Consider the function $V=ln(r+s^{2}+t^{3})$ Let us start differentiating the function with respect to $t$ keeping $r$ and $s$ constant. $\frac{∂V}{∂t}=\frac{∂}{∂t}[ln(r+s^{2}+t^{3})]$ $=\frac{1}{(r+s^{2}+t^{3})}\frac{∂}{∂t}[(r+s^{2}+t^{3})]$ $=\frac{3t^{2}}{(r+s^{2}+t^{3})}$ Differentiate with respect to $s$ keeping $r$ and $t$ constant. $\frac{∂^{2}V}{∂s∂t}=-\frac{3t^{2}}{(r+s^{2}+t^{3})^{2}}.2s$ (Apply power rule) $=-6st^{2}(r+s^{2}+t^{3})^{-2}$ Differentiate with respect to $r$ keeping $s$ and $t$ constant. $\frac{∂^{3}V}{∂r∂s∂t}=\frac{∂}{∂r}[\frac{∂^{2}}{∂s∂t}(-6st^{2}(r+s^{2}+t^{3})^{-2})]$ $=12st^{2}(r+s^{2}+t^{3})^{-3}$ Hence, $\frac{∂^{3}V}{∂r∂s∂t}=\frac{12st^{2}}{(r+s^{2}+t^{3})^{3}}$