Prob. 2 (b), Sec. 8.14, in Apostol's CALCULUS Vol 2: How to find the directional derivative of this scalar field?

Question

Let $f$ be the scalar field defined by the formula $$ f(x, y, z) = (x/y)^z, $$ for all points $(x, y, z) \in \mathbb{R}^3$ for which the formula makes sense.

Then what is the directional derivative of $f$ at the point $(1, 1, 1)$ in the direction of $2 \mathbf{i} + \mathbf{j} - \mathbf{k}$?

My Attempt:

Let us put $\mathbf{a} \colon= (1, 1, 1)$ and $\mathbf{y} \colon= (2, 1, -1)$. Then $\mathbf{a} + h \mathbf{y} = ( 1 + 2h, 1 + h, 1 - h)$, and so $$ \begin{align} \frac{ f( \mathbf{a} + h \mathbf{y} ) - f( \mathbf{a} ) }{ h } &= \frac{ [ (1+2h) / ( 1+h ) ]^{1-h} - 1 }{ h } \\ &= \end{align} $$

What next? How to find $$ \lim_{ h \to 0 } \frac{ f( \mathbf{a} + h \mathbf{y} ) - f( \mathbf{a} ) }{ h }? $$ Or, does this limit exist?

Answer 1

We have

• $f_x=\frac{z}{y}\left(\frac x y\right)^{z-1}$
• $f_y=-\frac{zx}{y^2}\left(\frac x y\right)^{z-1}$
• $f_z=\left(\frac x y\right)^{z}\log\left(\frac x y\right)$

then evaluate $\nabla f(1,1,1)$ and recall that the directional derivative along $v=2 \mathbf{i} + \mathbf{j} - \mathbf{k}$ is given by

$$f_v=\nabla f \cdot v$$