If a function$f(x,y,z)$ is differentiable at (a, b, c), explain

how to use the gradient$\nabla f(a,b,c)$to find the equation of

the hyperplane tangent to the graph of $f$at$(a,b,c)$.

Let $w=f(x,y,z)$is a function of two variables defined on an open set containing the point $(a,b,c)$and

Let $\Delta w=f(a+\Delta x,b+\Delta y,c+\Delta z)-f(a,b,c)$

If $f_x(a,b,c),f_y(a,b,c)$and $f_z(a,b,c)$exist, the function$f$ is said to be differentiable and

$\Delta w=f_x(a,b,c)\Delta x+f_y(a,b,c)\Delta y+f_z(a,b,c)\Delta z+{\epsilon }_1\Delta x+{\epsilon }_2\Delta y+{\epsilon }_3\Delta z\dots \dots \text{(1)}$

where ${\epsilon }_1,{\epsilon }_2$and ${\epsilon }_3$are functions of $\Delta x,\Delta y$and $\Delta z$and are zero when

$(\Delta x,\Delta y)\to (0,0)$

Substitute $\Delta x=x-a,\Delta y=y-b,\Delta z=z-c$and $\Delta w=f(x,y,z)-f(a,b,c)$

in $\left(1\right)$and eliminate ${\epsilon }_1\Delta x,{\epsilon }_2\Delta y$and ${\epsilon }_3\Delta z$terms.

$$\begin{gathered} ∆w=f_x(a,b,c)∆x+f_y(a,b,c)∆y+f_z(a,b,c)∆z+{\epsilon }_1∆x+{\epsilon }_2∆y+{\epsilon }_3∆z \\ f(x,y,z)-f(a,b,c)=f_x(a,b,c)(x-a)+f_y(a,b,c)(y-b)+f_z(a,b,c)(z-c) \\ w-f(a,b,c)=\left(f_x\right(a,b,c),f_y(a,b,c),f_z(a,b,c\left)\right)·\left(\right(x-a),(x-b),(x-c\left)\right) \\ =\nabla f(a,b,c)·\left(\right(x,y,z)-(a,b,c\left)\right) \end{gathered}$$

Hence, the equation of tangent plane to the surface $w=f(x,y,z)$at $(a,b,c)$is

$w-f(a,b,c)=\nabla f(a,b,c)·⟨(x,y,z)-(a,b,c)⟩$