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

I begin by writing the derivative as a dot product:

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

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

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

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

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

I begin by writing the derivative as a dot product:

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

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

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

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

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

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

CB