Directional Derivatives

Single-Variable Calculus Derivative

Recall that the definition of a single-variable calculus derivative is this $$\lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h} = f'(x)$$ What does this mean? When you move a very very small amount in the x-direction, what is the slope? As this increment gets smaller and smaller, you get the slope at that point.

Directional Derivative

The directional derivative at p in the direction v equals $$D_vf(p) = \lim_{t \to 0} \dfrac{f(p + tv) - f(p)}{t}$$ The magnitude of the vector has to be 1, otherwise we would get different answers in the same direction, which is undesirable.

Understanding this

Relating it back to $\mathbb{R}$, we get $$\lim_{t \to 0} \dfrac{f(p + t*1) - f(p)}{t}$$ More generally, imagine you're moving in the direction $<1,1>$.What is the slope in that specific direction? That is the directional derivative.

Partial Derivative

The partial derivative is a special case of the directional derivative. These are just directional derivatives in the direction of the basis vectors. They are written like this $$\dfrac{\partial f}{\partial x_n}(p) = D_{e_n}f(p)$$ These are easy to calculate. Take this for example $\dfrac{\partial f}{\partial x} x^2y + y^2x$. In this example, you could plug it into the directional derivative equation, which would be $$\lim_{t \to 0}\dfrac{(x+t)^2y + y^2(x+t) - x^2y - y^2x}{t} = \dfrac{x^2y - x^2y + 2txy + t^2y + xy^2 - xy^2 + ty^2}{t}$$ $$=\lim_{t \to 0}2xy + y^2 + ty = \boxed{2xy + y^2}$$ That was annoying. Now, how you would actually do it is consider y as a constant. For this example, we can rename $y$ to $c$ to make it clearer and do a normal derivative. $$\dfrac{d}{dx}x^2c + c^2x = 2xc + c^2 = 2xy + y^2$$

Computing Directional Derivatives

It turns out that we have this equation: $$D_vf(p) = Df(p)(v)$$ In other words, the directional derivative in the direction of v equals the total derivative at p of the vector. Therefore, we have an easy way of calculating this. $$D_vf(p) = Df(p)(v) = \begin{bmatrix}\dfrac{\partial f}{\partial x_1}(p) & \dfrac{\partial f}{\partial x_2}(p) & ... & \dfrac{\partial f}{\partial x_n}(p)\end{bmatrix}\begin{bmatrix}v_1 \\ v_2 \\ ... \\ v_n\end{bmatrix}$$ $$= \dfrac{\partial f}{\partial x_1}(p) * v_1 + \dfrac{\partial f}{\partial x_2}(p) * v_2 + ... + \dfrac{\partial f}{\partial x_n}(p) * v_n$$ Because partial derivatives are fairly easy to calculate, this means you don't have to use the limit definition whenever you want to find a directional derivative.

Examples: Partial Derivatives

$$\frac{\partial}{\partial x} e^{x^2 + cos(xy)}$$ Show/Hide

Start by looking at the exponent. $$=e^{x^2 + cos(xy)}\frac{\partial}{\partial x}\left(x^2 + cos(xy)\right)$$ $$=e^{x^2 + cos(xy)}\left(2x - ysin(xy)\right)$$

Examples: Directional Derivatives

$$f(x,y) = x^2 - 5xy \text{ in the direction of } <3, 4>$$ Show/Hide

First, what are the partials? $$\dfrac{\partial f}{\partial x} = 2x - 5y$$ $$\dfrac{\partial f}{\partial y} = -5x$$ Now, how can we turn the vector into a unit vector? Divide by its magnitude. $$<3, 4>*\frac{1}{\sqrt{3^2 + 4^2}} = <\frac{3}{5},\frac{4}{5}>$$ So, the directional derivative equals $$\frac{3}{5}\left(2x - 5y\right) +\frac{2}{5}\left(-5x\right)$$ $$=\dfrac{-4x}{5} - 3y$$