# Proof of Stokes’ Theorem Confirm the following step in the proof of Stokes’ Theorem. If z = s(x, y) and f, g, and h are functions of x, y, and z, with M = f + hz_x and N = g + hz_y, then M_y = ƒ_y + ƒ_z z_y + hz_(xy) + z_x(hy + h_z z_y) and N_x = g_x + g_z z_x + hz_(yx) + z_y(h_x + h_z z_x).

Proof of Stokes’ Theorem Confirm the following step in the proof of Stokes’ Theorem. If z = s(x, y) and f, g, and h are functions of x, y, and z, with M = f + hz_x and $N=g+h{z}_{y}$, then
${M}_{y}={ƒ}_{y}+{ƒ}_{z}{z}_{y}+h{z}_{xy}+{z}_{x}\left(hy+{h}_{z}{z}_{y}\right)$
and
${N}_{x}={g}_{x}+{g}_{z}{z}_{x}+h{z}_{yx}+{z}_{y}\left({h}_{x}+{h}_{z}{z}_{x}\right)$.
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Step 1
Considering the steps of proof of stokes theorem,
if, z = s(x,y)and f(x,y,z),g(x,y,z),h(x,y,z)
Hence the partial differentiation(using the chain rule) is,
$\frac{\partial M}{\partial y}=\frac{\partial \left(f\left(x,y,z\right)+h{z}_{x}\right)}{\partial y}$
$=\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}\frac{\partial z\left(x,y\right)}{dy}+h\frac{\partial {z}_{x}}{\partial y}+{z}_{x}\left(\frac{\partial h}{\partial y}\right)+\frac{\partial h}{\partial z}\frac{\partial z\left(x,y\right)}{\partial y}$
$={f}_{y}+{d}_{z}{z}_{y}+h{z}_{xy}+{z}_{x}\left({h}_{y}+{h}_{z}{z}_{y}\right)$
Step 2
Further,
$\frac{\partial N}{\partial x}=\frac{\partial \left(g\left(x,y,z\right)+h{z}_{y}\right)}{\partial x}$
$=\frac{\partial g}{\partial x}+\frac{\partial g}{\partial x}\frac{\partial z\left(x,y\right)}{\partial x}+h\frac{\partial {z}_{y}}{\partial x}+{z}_{y}\left(\frac{\partial h}{\partial x}+\frac{\partial h}{\partial z}\left(dz\frac{x,y}{dx}\right)\right)$
$={g}_{x}+{g}_{z}{z}_{x}+h{z}_{yx}+{z}_{y}\left({h}_{x}+{h}_{z}{z}_{x}\right)$