7. Compute the directional derivative of the function $g(x, y)=\sin (\pi(5 x-y))$ at the point $P(2,1)$ in the direction $\left\langle\frac{5}{13}, \frac{12}{13}\right\rangle$. 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 \(\nabla f = \left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right\rangle\).

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

Step 4 ：\(g = \sin(\pi*(5*x - y))\)

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

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

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 \(\left\langle\frac{5}{13}, \frac{12}{13}\right\rangle\).

Step 11 ：The directional derivative is \(-1.0\pi\)

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