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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?

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  • $\begingroup$ I think you need to find $\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}$ at $(1,1,1)$, and then find the dot product of this vector with $(2,1,-1)$. $\endgroup$ – fierydemon Jun 13 '18 at 12:48
  • $\begingroup$ @fierydemon does this formula hold in every situation where the partials exist? According to the text by Apostol, this works only when $f$ is differentiable at $\mathbf{a}$. $\endgroup$ – Saaqib Mahmood Jun 13 '18 at 13:01
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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

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

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  • $\begingroup$ does the formula work in all cases when the partial derivatives exist? I thought it works only when the scalar field is differentiable. $\endgroup$ – Saaqib Mahmood Jun 13 '18 at 13:03
  • $\begingroup$ @SaaqibMahmood Yes exactly we need that f is differentiable at that point $\endgroup$ – gimusi Jun 13 '18 at 13:35
  • $\begingroup$ how then do we know that this particular $f$ is differentiable, and at what points $(x, y, z)$ in $\mathbb{R}^3$? $\endgroup$ – Saaqib Mahmood Jun 13 '18 at 14:18
  • $\begingroup$ At the point (1,1,1) the function is differentiable since the partial derivatives exist and are continuous at that point. $\endgroup$ – gimusi Jun 13 '18 at 22:23

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