# Thread: Gradient Vector max/min

1. ## Gradient Vector max/min

I need to show that a differentiable function $f(x_1,x_2...x_n)$ decreases most rapidly at $X$ in the direction opposite to the gradient vector, that is, in the direction of $-\bigtriangledown f(X)$

I begin by writing the derivative as a dot product:

$D_u f(X)=u\cdot\bigtriangledown f(X)$

$=\mid u\bigtriangledown f(X)\mid \cos\theta_{u,f}$

$\cos\theta_{u,f}$ is smallest when $\theta_{u,f}=\pi+2n\pi$ i.e when $\cos\theta_{u,f}=-1$. The derivative $D_u f(X)$ becomes $-\mid\bigtriangledown f(X)\mid=-\bigtriangledown f(X)$.

Have I shown that the minimum of a directional derivative at a point $X$ is the negative of the gradient? Is this what they're asking for?

2. Originally Posted by adkinsjr
I need to show that a differentiable function $f(x_1,x_2...x_n)$ decreases most rapidly at $X$ in the direction opposite to the gradient vector, that is, in the direction of $-\bigtriangledown f(X)$

I begin by writing the derivative as a dot product:

$D_u f(X)=u\cdot\bigtriangledown f(X)$

$=\mid u\bigtriangledown f(X)\mid \cos\theta_{u,f}$

$\cos\theta_{u,f}$ is smallest when $\theta_{u,f}=\pi+2n\pi$ i.e when $\cos\theta_{u,f}=-1$. The derivative $D_u f(X)$ becomes $-\mid\bigtriangledown f(X)\mid=-\bigtriangledown f(X)$.

Have I shown that the minimum of a directional derivative at a point $X$ is the negative of the gradient? Is this what they're asking for?
$u$ is a unit vector,

$D_u f(x)=u. \nabla f (X)=|\nabla f(X)| \cos(\theta_{u,f})$

which has a minimum when $\cos(\theta_{u,f})=-1$ which is when $u$ and $\nabla f(X)$ point in oposite directions.

CB