Q:

directional derivative problem

for a point M(4,1) and a function $z  = x y^2 - (x^2/y^3)$ I was tasked with finding a directional derivative in the direction which creates a 30 degree angle with the $x$ axis....I find it a little hard to imagine and would like to get help on how this might be done..

A:

The directional derivative of a scalar function
$f(\mathbf{x}) = f(x_1, x_2, \ldots, x_n)$
along a vector
$\mathbf{v}$ is, by definition
$$\nabla_{\mathbf{v}}{f}(\mathbf{x}) = \lim_{h \rightarrow 0}{\frac{f(\mathbf{x} + h\mathbf{v}) - f(\mathbf{x})}{h}}\hbox{.}$$
In case when $\nabla{f}(\mathbf{x})$ exists, $\nabla_{\mathbf{v}}{f}(\mathbf{x}) = \nabla f(\mathbf{x}) \cdot \mathbf{v}$ if $|\mathbf{v}|=1$
where the $\nabla$ denotes the gradient and $\cdot$ is the dot product.
So one just takes the gradient (it exists in most cases); in the OP
$$\hbox{grad}\,z = \left(\frac{\partial}{\partial x}(x∗y^2−(x^2/y^3));
\frac{\partial}{\partial y}(x∗y^2−(x^2/y^3))\right)=\left(y^2-2x/y^3;2xy+3x^2/y^4\right),$$
takes the vector of norm $1$ pointing in given direction and multiplies them.
Substituting concrete variables, if necessary.
The result in the OP would be $(\hbox{grad}\, z).(\frac{\sqrt{3}}{2};\frac{1}{2})$ , the rest's up to you.

