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Let $f$ be real-valued and differentiable at a point $c$ in $\mathbb R^n$, and assume that $||\nabla f(c)|| \neq 0.$ Prove that there is one and only one unit vector $u$ in $R^n$ such that $|f(c; u)|$ = $||\nabla f(c)||$ and that this is the unit vector for which $|f(c; u)|$ has its maximum value.

I can't think of any way to solve this, any hints are appreciated.

$f(c;u)$ is directional derivative of $f$ at $c$ in vector $u$'s direction

john doe
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  • See https://math.stackexchange.com/q/1912660/265466 for several approaches. – amd Sep 20 '17 at 20:54
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    $f(c;u) = \nabla f \cdot u = \left| \nabla f \right| \cdot \cos(\theta)$, where $\theta$ is the angle between $\nabla f$ and $u$. So obviously it is maximized when $\theta = 0$, which happens when $u$ points in the same direction as $\nabla f$. – Nick Sep 22 '17 at 16:42

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