# Is this derivation valid?

1. Jun 28, 2013

### ayao

For directional derivatives:

Let $\hat{u}=<a,b,c>$ be the direction.

Thus, $\frac{∂\hat{u}}{∂x}=\frac{\sqrt{a^2+b^2+c^2}}{a}$ and so on. So,

$\frac{∂x}{∂\hat{u}}=\frac{a}{\sqrt{a^2+b^2+c^2}}=a$

Thus,

$\frac{∂F}{∂\hat{u}}=\frac{∂F}{∂x}a+\frac{∂F}{∂y}b+\frac{∂F}{∂z}c=∇F \bullet \hat{u}$.

Last edited: Jun 28, 2013
2. Jun 28, 2013

### Staff: Mentor

Are a, b, and c constants? If so, the derivative of $\hat{u}$ would be the zero vector.

3. Jun 28, 2013

### ayao

I probably should have defined it better; $\hat{u}$ is a unit vector in the direction that we are trying to find the derivative in. The linear derivatives of $\hat{u}$ are defined as derivatives of a line in that direction.