Angle between functions

For any function $h$ such that $h'$ exists at $x_{0}$ , the tangent vector to the graph at the point $\left ( x_{0}, h(x_{0}) \right )$ may be found in this way:

$h'(x_{0}) = \frac{h_{1}}{h_{2}} \;\;\; \Rightarrow \overrightarrow{\mathrm{v}}_{h, x_{0}} = \left ( h_{1}, h_{2} \right ) = h_{2} \left ( h'(x_{0}), 1 \right )$

where $\overrightarrow{\mathrm{v}}_{h, x_{0}}$ is the tangent vector. Find the unitary vector in the same direction as follows:

$\overrightarrow{\mathrm{u}}_{h, x_{0}} = \frac{\overrightarrow{\mathrm{v}}_{h, x_{0}}}{\left \| \overrightarrow{\mathrm{v}}_{h, x_{0}} \right \|} = \frac{\left ( h'(x_{0}), 1 \right )}{\sqrt{h'(x_{0})^{2} + 1}}$

This is valid both for $f$ and $g$ . The inner product between $\overrightarrow{\mathrm{u}}_{f, x_{0}}$ and $\overrightarrow{\mathrm{u}}_{g, x_{0}}$ will give us, of course, the cosine of the angle we're looking for:

$\frac{f'(x_{0})g'(x_{0})+1}{\sqrt{\left ( f'(x_{0})g'(x_{0}) \right )^{2} + \left ( f'(x_{0}) \right )^{2} + \left ( g'(x_{0}) \right )^{2} + 1}} = \cos \left \langle f, g \right \rangle_{x_{0}}$

Now, if we let $f(x) = \ln(x)$ and $g(x) = x^{2} - x$ , we have $x_{0} = 1$ . All we need are the derivatives of those functions evaluated at $x = 1$ , then. It is easy to show that both of these values are 1. Then:

$\cos \left \langle \ln(x), x^{2} - x \right \rangle_{1} = 1$

Therefore, $\left \langle \ln(x), x^{2} - x \right \rangle_{1} = 0$ .