Given the function
$f(x,y)=g(2x+y), g'(3)=3$
Find the coordinates of a unit vector $v=(v_1,v_2)$ that finds the maximum directional derivative $D_ uf(1,1)$ of f in the point $(1,1)$.
I know that I should find $\nabla f(x,y)*u$ in such a way that u maximizes it. But I am uncertain how to do so.
$\endgroup$ 22 Answers
$\begingroup$We have given $y=g(t),\, \dfrac{dg}{dt}=g^\prime(t),\, g^\prime(3)=3,\,t=2x+y,\,f(x,y)=g(2x+y)$
Therefore, by the chain rule, $f_x(x,y)=\dfrac{dg}{dt}\cdot\dfrac{\partial t}{\partial x}=2g^\prime(t)$ and $f_y(x,y)=\dfrac{dg}{dt}\cdot\dfrac{\partial t}{\partial y}=g^\prime(t)$.
So $\nabla f(1,1)=\langle2,1\rangle g^\prime(t)=\langle2,1\rangle g^\prime(3)=\langle6,3\rangle$
$\vert \nabla f(1,1)\vert=3\sqrt{5}$, therefore $u=\dfrac{\nabla f(1,1)}{\vert \nabla f(1,1)\vert}=\left\langle \dfrac{2\sqrt{5}}{5},\dfrac{\sqrt{5}}{5}\right\rangle$.
$\endgroup$ 2 $\begingroup$note that the partial derivatives of $f(x,y)$ are: $$ f_x=g' \frac{\partial}{\partial x}(2x+y)=2g' \qquad f_y=g' \frac{\partial}{\partial y}(2x+y)=g' $$ so the gradient at $(1,1)$ is: $$ \nabla f(x,y)= (2g'(3),g'(3))^T=(6,3)^T $$
and, as noted in the comment, the directional derivative is maximized for a vector parallel to the gradient.
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