1. **State the problem:** Find the gradient of the function $f(x,y) = \cos(x^2 + y^2)$ at the point $(3, -4)$. 2. **Recall the gradient formula:** The gradient of a function $f(x,y)$ is given by $$\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right).$$ 3. **Calculate partial derivatives:** - For $\frac{\partial f}{\partial x}$, use the chain rule: $$\frac{\partial f}{\partial x} = -\sin(x^2 + y^2) \cdot \frac{\partial}{\partial x}(x^2 + y^2) = -\sin(x^2 + y^2) \cdot 2x.$$ - For $\frac{\partial f}{\partial y}$, similarly: $$\frac{\partial f}{\partial y} = -\sin(x^2 + y^2) \cdot \frac{\partial}{\partial y}(x^2 + y^2) = -\sin(x^2 + y^2) \cdot 2y.$$ 4. **Write the gradient vector:** $$\nabla f = \left(-2x \sin(x^2 + y^2), -2y \sin(x^2 + y^2)\right).$$ 5. **Evaluate at the point $(3, -4)$:** - Compute $x^2 + y^2 = 3^2 + (-4)^2 = 9 + 16 = 25$. - Compute $\sin(25)$ (in radians). 6. **Substitute values:** $$\nabla f(3,-4) = \left(-2 \cdot 3 \cdot \sin(25), -2 \cdot (-4) \cdot \sin(25)\right) = \left(-6 \sin(25), 8 \sin(25)\right).$$ **Final answer:** $$\boxed{\nabla f(3,-4) = \left(-6 \sin(25), 8 \sin(25)\right)}.$$