7. Compute the directional derivative of the function $g(x,y)=\sin (\pi (5x-y))$ at the point $P(2,1)$ in the direction $⟨\frac{5}{13},\frac{12}{13}⟩$. Be sure to use a unit vector for the direction vector.
The directional derivative is
(Type an exact answer, using $\pi$ as needed.)

Step 1 ：The directional derivative of a function $f$ at a point $P$ in the direction of a unit vector $\mathbf{u}$ is given by the dot product of the gradient of $f$ at $P$ and $\mathbf{u}$.

Step 2 ：The gradient of a function $f(x,y)$ is given by $\mathrm{\nabla }f=⟨\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}⟩$.

Step 3 ：We first need to compute the partial derivatives of $g(x,y)$ with respect to $x$ and $y$.

Step 4 ：$g=\sin (\pi \ast (5\ast x-y))$

Step 5 ：$\frac{\partial g}{\partial x}=5\pi \cos (\pi \ast (5\ast x-y))$

Step 6 ：$\frac{\partial g}{\partial y}=-\pi \cos (\pi \ast (5\ast x-y))$

Step 7 ：We then evaluate these at the point $P(2,1)$.

Step 8 ：$\frac{\partial g}{\partial x}$ at $P$ is $-5\pi$

Step 9 ：$\frac{\partial g}{\partial y}$ at $P$ is $\pi$

Step 10 ：Finally, we take the dot product of the resulting vector with the given direction vector $⟨\frac{5}{13},\frac{12}{13}⟩$.

Step 11 ：The directional derivative is $-1.0\pi$

Step 12 ：The directional derivative of the function $g(x,y)=\sin (\pi (5x-y))$ at the point $P(2,1)$ in the direction $⟨\frac{5}{13},\frac{12}{13}⟩$ is $\boxed{{\displaystyle -\pi }}$