13. Find the first partial derivatives of the following function. \[ \begin{array}{l} g(x, y)=\cos ^{5}\left(x^{6} y^{7}\right) \\ \frac{\partial g}{\partial x}=\square \\ \frac{\partial g}{\partial y}=\square \end{array} \]

Step 1 ：First, we need to understand the meaning of the question. We are asked to find the first partial derivatives of the function \(g(x, y)=\cos ^{5}\left(x^{6} y^{7}\right)\).

Step 2 ：Let's start with the partial derivative with respect to \(x\). We will use the chain rule, which states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.

Step 3 ：Applying the chain rule, we get \(\frac{\partial g}{\partial x} = 5\cos^{4}(x^{6}y^{7})\cdot(-\sin(x^{6}y^{7}))\cdot(6x^{5}y^{7})\).

Step 4 ：Simplifying, we get \(\frac{\partial g}{\partial x} = -30x^{5}y^{7}\cos^{4}(x^{6}y^{7})\sin(x^{6}y^{7})\).

Step 5 ：Now, let's find the partial derivative with respect to \(y\). Again, we will use the chain rule.

Step 6 ：Applying the chain rule, we get \(\frac{\partial g}{\partial y} = 5\cos^{4}(x^{6}y^{7})\cdot(-\sin(x^{6}y^{7}))\cdot(7x^{6}y^{6})\).

Step 7 ：Simplifying, we get \(\frac{\partial g}{\partial y} = -35x^{6}y^{6}\cos^{4}(x^{6}y^{7})\sin(x^{6}y^{7})\).

Step 8 ：So, the first partial derivatives of the function \(g(x, y)=\cos ^{5}\left(x^{6} y^{7}\right)\) are \(\frac{\partial g}{\partial x} = -30x^{5}y^{7}\cos^{4}(x^{6}y^{7})\sin(x^{6}y^{7})\) and \(\frac{\partial g}{\partial y} = -35x^{6}y^{6}\cos^{4}(x^{6}y^{7})\sin(x^{6}y^{7})\).

Step 9 ：Finally, we check our results. The derivatives are correct if they satisfy the definition of a derivative and the rules of differentiation. In this case, they do, so our final answer is \(\boxed{\frac{\partial g}{\partial x} = -30x^{5}y^{7}\cos^{4}(x^{6}y^{7})\sin(x^{6}y^{7})}\) and \(\boxed{\frac{\partial g}{\partial y} = -35x^{6}y^{6}\cos^{4}(x^{6}y^{7})\sin(x^{6}y^{7})}\).