# Simple directional derivative question

I'm having some trouble solving this directional derivative problem.

$$f(x,y) = \sin(2x + 5y), P(-15, 6), \mathbf{u} = {1 \over 2} (\sqrt{3}\mathbf{i-j})$$

I know the theorem for the bivariate directional derivative: $$D_u f(x, y) = f_x(x, y)a + f_y(x, y)b$$

But that is with the unit vector in form $\mathbf{u}=\langle a,b\rangle$. I think perhaps I am not translating the unit vector into the proper format.

I got the answer: $$\sqrt{3} \mathbf{i}\cos(2x+5y)+\cos(2x+5y){-5 \over 2}\mathbf{j}$$ But that is incorrect.

• Notationally, $\langle a,b\rangle=a\mathbf{i}+b\mathbf{j}$. So in this case $\mathbf{u}=\frac{1}{2}(\sqrt{3}\mathbf{i}-\mathbf{j})=\langle\frac{\sqrt{3}}{2},-\frac{1}{2}\rangle$ – charlestoncrabb Dec 8 '15 at 23:25
• Also you need to plug the point $P$ into your cosines... – charlestoncrabb Dec 8 '15 at 23:26
• @charlestoncrabb Thanks, got it. – d0rmLife Dec 8 '15 at 23:37

$$\sqrt{3} - {5 \over 2 }$$