Diffrentiability in Multivariate calculus Let,$f(x, y, z) =x^3+y^3+z^3$
$L$ be a linear map from $\mathbb R^3$ to $\mathbb R$
Satisfying
$$\displaystyle\lim_{(x, y, z) \to (0, 0,0)} \frac{f(1+x, 1+y, 1+z) -f(1, 1,1) -L(x, y, z) }{\sqrt{x^2+y^2+z^2}}=0$$
Then find the value of  $L(1, 2,4)$
I am unable to understand how to approach. Pls help
 A: Your problem describes the Frechet Derivative of the function $f(x,y,z) = x^3 + y^3 + z^3$ at the point $(1,1,1)$ evaluated in the direction $(1,2,4)$.
Here is how to do this from first principles.  If such a map $L$ exists, then we have
$$
\begin{align*}
0 &= \lim_{(h,0,0) \to 0} \frac{f(1+h,1,1) - f(1,1,1) - L(h,0,0)}{h}\\
 &= \lim_{h \to 0} \frac{f(1+h,1,1) - f(1,1,1)}{h} - \frac{L(h,0,0)}{h}\\
 &= \lim_{h \to 0} \frac{(1+h)^3 + 1 + 1 - (1+1+1)}{h} - \frac{hL(1,0,0)}{h}\\
 &= \lim_{h \to 0} \frac{3h+3h^2+h^3}{h} - L(1,0,0)\\
 &= 3 - L(1,0,0)
\end{align*}$$
So $L(1,0,0) = 3$.  Note that this is exactly $\frac{\partial f}{\partial x} \big|_{(1,1,1)}$.
Similarly, by symmetry, we have $L(0,1,0) = L(0,0,1) = 3$.
Thus $L$ must be the linear map with the matrix $\begin{bmatrix} 3 & 3 & 3\end{bmatrix}$ if it exists.  To show that $L$ actually does satisfy the definition you need to go back to the original limit and verify that it actually does equal $0$.
Once you very this,  $L(1,2,4) = \begin{bmatrix} 3 & 3 & 3\end{bmatrix} \begin{bmatrix}  1 \\ 2 \\ 4\end{bmatrix} = 3(1) + 2(3) + 4(3) = 21$.
A: Hint:
For the 1st part the gradient
$$L(1,1,1)=
\left[\frac{\partial f}{\partial x}|_{(1,1,1)}\qquad \frac{\partial f}{\partial y}|_{(1,1,1)}\qquad \frac{\partial f}{\partial z}|_{(1,1,1)} \right],$$
will do the task.
A: The definition of the linear map is rather strange ? I will go this way:
Define the vector
$\mathbf{x}_0 = (1,1,1)$ and
$\mathbf{v} 
= (x,y,z)$.
Let $t=\| \mathbf{v} \|$,
and define the unit vector
$\mathbf{w}=
\frac{\mathbf{v}}{\|\mathbf{v}\|}$
Thus the directional derivative writes
$$
Df(\mathbf{x}_0)[\mathbf{w}]
=
\lim_{t\to 0}
\frac{f(\mathbf{x}_0+t\mathbf{w})-f(\mathbf{x}_0)}{t}
=
\nabla f(\mathbf{x}_0):\mathbf{w}
$$
where the colon operator stands for the inner product between two vectors.
I think you are asked to compute the directional derivative in the
direction
$\mathbf{w}=
\frac{\mathbf{v}}{\|\mathbf{v}\|}$
with $\mathbf{v}=(1,2,4)$
To do so, you need to compute the gradient
$\nabla f(\mathbf{x}_0)=3(x_0^2,y_0^2,z_0^2)
=3(1,1,1)$.
Finally
$$
3(1,1,1):\frac{1}{\sqrt{261}}(1,2,4)
=
\frac{21}{\sqrt{261}}
$$
UPDATE :
Define the vector
$\mathbf{x}_0 = (1,1,1)$ and
$\mathbf{v} 
= (x,y,z)$.
Define the unit vector
$\mathbf{w}=
\frac{\mathbf{v}}{t}$
with $t=\| \mathbf{v} \|$.
The directional derivative at point $\mathbf{x}_0$
in the direction $\mathbf{w}$
writes
$$
Df(\mathbf{x}_0)[\mathbf{w}]
=
\lim_{t\to 0}
\frac{f(\mathbf{x}_0+t\mathbf{w})-f(\mathbf{x}_0)}{t}
$$
We can write
$$
\lim_{t\to 0}
\frac{f(\mathbf{x}_0+t\mathbf{w})-f(\mathbf{x}_0)-
t Df(\mathbf{x}_0)[\mathbf{w}]}{t}
=0
$$
Because
$t Df(\mathbf{x}_0)[\mathbf{w}]
= Df(\mathbf{x}_0)[t\mathbf{w}]$,
it holds
$$
\lim_{t\to 0}
\frac{f(\mathbf{x}_0+t\mathbf{w})-f(\mathbf{x}_0)-
Df(\mathbf{x}_0)[t\mathbf{w}]}{t}
=0
$$
or equivalently
$$
\lim_{\| \mathbf{v} \|\to 0}
\frac{f(\mathbf{x}_0+\mathbf{v})-f(\mathbf{x}_0)-
Df(\mathbf{x}_0)[\mathbf{v}]}{\| \mathbf{v} \|}
=0
$$
We deduce that
$L(\mathbf{v})
=Df(\mathbf{x}_0)[\mathbf{v}]
=\nabla f(\mathbf{x}_0):\mathbf{v}
$
and thus the correct result is 21.
