Let f(x, y) = (x° + ay´, xʻy°, 2x+by), g(u,v,w) = (3u² + bv² + w°, uvw). Find the derivative of f(g(u, v, w)) using the chain rule.

$$\begin{gathered} \mathrm{f}(\mathrm{x},\mathrm{y})=({\mathrm{x}}^3+{\mathrm{ay}}^2,{\mathrm{x}}^2{\mathrm{y}}^3,2\mathrm{x}+\mathrm{by}) \\ \mathrm{g}(\mathrm{u},\mathrm{v},\mathrm{w})=(3{\mathrm{u}}^2+{\mathrm{bv}}^2+{\mathrm{w}}^3,\mathrm{uvw}) \\ \mathrm{To} \mathrm{Find}: \mathrm{Derivative} \mathrm{of} \mathrm{f}\left(\mathrm{g}\right(\mathrm{u},\mathrm{v},\mathrm{w}\left)\right) \mathrm{using} \mathrm{the} \mathrm{chain} \mathrm{rule}. \end{gathered}$$

$$\begin{gathered} \mathrm{We} \mathrm{use} \mathrm{the} \mathrm{chain} \mathrm{rule} \mathrm{to} \mathrm{differentiate} \mathrm{composite} \mathrm{function}: \\ \mathrm{If} \mathrm{f} \mathrm{and} \mathrm{g} \mathrm{are} \mathrm{both} \mathrm{differentiable} \mathrm{and} \mathrm{F}=\mathrm{f}°\mathrm{g}(\mathrm{i}.\mathrm{e}.\mathrm{F}(\mathrm{x})=\mathrm{f}(\mathrm{g}\left(\mathrm{x}\right)\left)\right) \mathrm{is} \mathrm{the} \mathrm{composite} \mathrm{function}, \\ \mathrm{then} \mathrm{F}\text{'}\left(\mathrm{x}\right)=\mathrm{f}\text{'}\left(\mathrm{g}\right(\mathrm{x}\left)\right)\bullet \mathrm{g}\text{'}\left(\mathrm{x}\right) \\ \mathrm{Now}, \mathrm{derivative} \mathrm{of} \mathrm{a} \mathrm{multivariable} \mathrm{vector} \mathrm{function} \mathrm{is} \mathrm{given} \mathrm{as}: \\ \frac{\partial \mathrm{F}}{\partial {\mathrm{x}}_{\mathrm{i}}}=⟨\frac{\partial {\mathrm{F}}_1}{\partial {\mathrm{x}}_{\mathrm{i}}},\frac{\partial {\mathrm{F}}_2}{\partial {\mathrm{x}}_{\mathrm{i}}},\dots ,\frac{\partial {\mathrm{F}}_{\mathrm{m}}}{\partial {\mathrm{x}}_{\mathrm{i}}}⟩ \end{gathered}$$

$$\begin{gathered} \mathrm{F}=\mathrm{f}\circ \mathrm{g} \\ =\mathrm{f}\left(\mathrm{g}\right(\mathrm{u},\mathrm{v},\mathrm{w}\left)\right) \\ =\mathrm{f}\left(\right(3{\mathrm{u}}^2+{\mathrm{bv}}^2+{\mathrm{w}}^3,\mathrm{uvw}\left)\right) \\ =\left(\right(3{\mathrm{u}}^2+{\mathrm{bv}}^2+{\mathrm{w}}^3{)}^3+\mathrm{a}(\mathrm{uvw}{)}^2,(3{\mathrm{u}}^2+{\mathrm{bv}}^2+{\mathrm{w}}^3{)}^2(\mathrm{uvw}{)}^3,2(3{\mathrm{u}}^2+{\mathrm{bv}}^2+{\mathrm{w}}^3)+\mathrm{b}(\mathrm{uvw}\left)\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Next}, \mathrm{we} \mathrm{find} \mathrm{the} \mathrm{derivative} \mathrm{of} \mathrm{the} \mathrm{f}\left(\mathrm{g}\right(\mathrm{u},\mathrm{v},\mathrm{w}\left)\right) \mathrm{using} \mathrm{the} \mathrm{chain} \mathrm{rule}. \\ \mathrm{Let} {\mathrm{F}}_1=(3{\mathrm{u}}^2+{\mathrm{bv}}^2+{\mathrm{w}}^3{)}^3+\mathrm{a}(\mathrm{uvw}{)}^2, {\mathrm{F}}_2=(3{\mathrm{u}}^2+{\mathrm{bv}}^2+{\mathrm{w}}^3{)}^2(\mathrm{uvw}{)}^3, {\mathrm{F}}_3=2(3{\mathrm{u}}^2+{\mathrm{bv}}^2+{\mathrm{w}}^3)+\mathrm{b}\left(\mathrm{uvw}\right) \\ \frac{\partial \mathrm{F}}{\partial \mathrm{u}}=⟨\frac{\partial {\mathrm{F}}_1}{\partial \mathrm{u}},\frac{\partial {\mathrm{F}}_2}{\partial \mathrm{u}},\frac{\partial {\mathrm{F}}_3}{\partial \mathrm{u}}⟩ \\ \frac{\partial \mathrm{F}}{\partial \mathrm{v}}=⟨\frac{\partial {\mathrm{F}}_1}{\partial \mathrm{v}},\frac{\partial {\mathrm{F}}_2}{\partial \mathrm{v}},\frac{\partial {\mathrm{F}}_3}{\partial \mathrm{v}}⟩ \\ \frac{\partial \mathrm{F}}{\partial \mathrm{w}}=⟨\frac{\partial {\mathrm{F}}_1}{\partial \mathrm{w}},\frac{\partial {\mathrm{F}}_2}{\partial \mathrm{w}},\frac{\partial {\mathrm{F}}_3}{\partial \mathrm{w}}⟩ \end{gathered}$$