# computing the directional derivative by definition

The definition of directional derivative:

\begin{equation} \frac{\partial f(\mathbf{x})}{\partial \mathbf{v}} = \lim_{h \rightarrow 0} \frac{f(\mathbf{x} + h \mathbf{v})-f(\mathbf{x})}{h}. \end{equation}

I have a function: $$F(x,y) = 2x^4 + 3y^3 +5xy$$ And the questin is: How does the function change, when we starting from point $(2,3)$ and the value of $x$ decreases on $0.067$ and the value of $y$ increases on $0.033$.

I think, that in order to solve this problem we need to compute the directional derivative by definition. So as i understood we just plug $(2,3)$ in $x$ and $(-0.067,0.033)$ is our direction vector $v$?Maybe i am wrong.

How should it looks like?

Edited:

As i understood we also need to convert our direction vector $(-0.067,0.033)$ to the unit vector. But if i am right in all my suggestions it would be terrible hard to solve this problem by definition(because of 4 and 3 power of the function).

Maybe there are other ways of solving thins problem?

• Solving it by using the definition directly isn’t that hard, just a bit tedious. Remember that $\mathbf v$ is a fixed vector, so you’ll just be expanding the polynomial $f(x+hv_x,y+hv_y)-f(x,y)$, with $v_x$ and $v_y$ constants. However, you can immediately drop any term that’s quadratic or greater in $h$ since it will go to $0$ when you take the limit, so you only have to compute a few terms of the expansion.
– amd
Jun 23 '16 at 15:13

Your function is a polynomial, hence differentiable. For a differentiable function $f$ the directional derivative in direction $v$ at point $x_0$ may be computed as follows: Compute the gradient of $f$ at $x_0$, and take the inner product with $v$. In your case this yields: $$\nabla f(2,3) = (79, 91)$$ $$(79,91)\cdot (-0.067,0.033) = -2.29$$