Directional derivative of function at specific point

Question

If I've a function (in this case also differentiable):

x^3/3 - x*y^2 - x/8 + y^2

how can I find quickly the directional derivative of the function in (1,1) with respect to the unit vector v(sqrt(3)/2, 1/2). Finding the gradient, and run the dot product. But what if the function had not been differentiable?

Answer 1

Here is an example, where a function of two variables

f[x_,y_]:=Sqrt[RealAbs[x^2 - x*y]]

is not differentiable at the origin, but its directional derivative at the origin for many directions exists.

We consider the restriction of the function f[x,y] on the ray from the origin along a vector {a,b}, preserving the scale by

g[t] := f[x, y] /. {x -> a/Norm[{a, b}]*t, y -> b/Norm[{a, b}]*t}

assuming t>=0. Now

D[g[t],t]
(*(((2 a^2 t)/(Abs[a]^2 + Abs[b]^2) - (2 a b t)/(   Abs[a]^2 + Abs[b]^2)) ((a^2 t^2)/
(Abs[a]^2 + Abs[b]^2) - (a b t^2)/(Abs[a]^2 + Abs[b]^2)))/(2 RealAbs[(a^2 t^2)/
( Abs[a]^2 + Abs[b]^2) - (a b t^2)/(Abs[a]^2 + Abs[b]^2)]^(3/2))*)

and

Limit[%, t -> 0, Direction -> "FromAbove"] // Simplify

$$ \begin{cases} \sqrt{-\frac{a (a-b)}{a^2+b^2}} & a\neq 0\land a^2<a b \\ \sqrt{\frac{a (a-b)}{a^2+b^2}} & (b=0\land a\neq 0)\lor (b<0\land (a\geq 0\lor a\leq b))\lor (b>0\land (a\geq b\lor a\leq 0)) \\ \text{Indeterminate} & \text{True} \\ \end{cases} $$ It should be noticed that

ResourceFunction["DirectionalD"][Sqrt[RealAbs[x^2 - x*y]], {a, b}, {x, y}]
(*-((b x (x^2 - x y))/(2 RealAbs[x^2 - x y]^(3/2))) +(a (2 x - y) (x^2 - x y))/(2 RealAbs[x^2 - x y]^(3/2))*)

, but

Limit[%, {x, y} -> {0, 0}]
(*Indeterminate*)

Answer 2

Use the definition of the directional derivative: Let f,z,v be vectors and t a scalar, the directional derivative of f[z] along v is

Limit[(f[z + t v]-f[z])/t,t->0]

For the above example with f[x,y]=Sqrt[RealsAbs[x^2-x y]]

Limit[((f[x, y] /. {Thread[{x, y} -> {x, y} + t {1, 1/2}]}) - 
    f[x, y])/t, t -> 0, Direction -> "FromAbove"]