## Calculus 8th Edition

$\frac{∂^{6}u}{∂x∂y^{2}∂z^{3}}=abc(b-1)(c-1)(c-2)x^{a-1}y^{b-2}z^{c-3}$
Consider the function $u=x^{a}y^{b}z^{c}$ Let us start by differentiating the function partially thrice times with respect to $z$ keeping $x$ and $y$ constant . $\frac{∂u}{∂z}=\frac{∂}{∂z}[x^{a}y^{b}z^{c}]=cx^{a}y^{b}z^{c-1}$ $\frac{∂^{2}u}{∂z^{2}}=\frac{∂}{∂z}[cx^{a}y^{b}z^{c-1}]=c(c-1)x^{a}y^{b}z^{c-2}$ $\frac{∂^{3}u}{∂z^{3}}=\frac{∂}{∂z}[c(c-1)x^{a}y^{b}z^{c-2}]=c(c-1)(c-2)x^{a}y^{b}z^{c-3}$ Now, differentiating $\frac{∂^{3}u}{∂z^{3}}$ partially twice times with respect to $y$ keeping $x$ and $z$ constant . $\frac{∂^{4}u}{∂y∂z^{3}}=bc(c-1)(c-2)x^{a}y^{b-1}z^{c-3}$ and $\frac{∂^{5}u}{∂y^{2}∂z^{3}}=bc(b-1)(c-1)(c-2)x^{a}y^{b-2}z^{c-3}$ Again differentiating $\frac{∂^{5}u}{∂y^{2}∂z^{3}}$ with respect to $x$ keeping $y$ and $z$ constant. Hence, $\frac{∂^{6}u}{∂x∂y^{2}∂z^{3}}=abc(b-1)(c-1)(c-2)x^{a-1}y^{b-2}z^{c-3}$