# Maximum directional derivative

At a certain point $P$ the function $f(x,y)$ has a direction derivative $\sqrt{2}$ in the direction of $\hat i-\hat j$ and $3\sqrt{2}$ in the direction of $\hat i+\hat j$. What is the maximum directional derivative of $f$ at the point $P$

From the first directional derivative we got the equation $\frac{f_x}{\sqrt{2}} -\frac{f_y }{\sqrt{2}} =\sqrt{2}\rightarrow f_x -f_y=2$ From the second $f_x+f_y=6$ hence $f_x=4 , f_y=2$ now we want to maximize $(4,2).\frac{\hat v}{|| \hat v||}$ how to continue? Is there any easier approach?

Yes, the directional derivative is maximal in the direction pointing along the gradient, i.e., $$\hat v = \frac{1}{\|\nabla f\|}\nabla f\,.$$ This is a general result of multivariable calculus. Since you know the partial derivatives, it is easy to compute $\nabla f$ and $\hat v$.
• $(4i+2j).\frac{4i+2j}{\sqrt{20}}$ is the answer? Thanks i got it – IrbidMath Mar 9 '17 at 16:43