\[ f(x, y, x)=\left(x y^{2} k, e^{x y}+x+x y x\right), \quad f(x, y, x)=\ln \left(x+x^{2}+y^{2}+x^{2}\right) \] Qalculafl divat, rot it yi grad \( f \). \[ \operatorname{grad}(r \varphi)=r \operatorname{grad} \psi+\frac{r}{r} p \]

To calculate the gradient of the function \( f(x, y, x) = \ln \left(x + x^2 + y^2 + x^2\right) \), we first simplify the function to \( f(x, y) = \ln \left( 2x^2 + y^2 + x \right) \). The gradient, denoted \( \nabla f \), can be computed by finding the partial derivatives with respect to \( x \) and \( y \). The real-world application of gradients lies in optimization problems—think of how companies use them to find maximum profit or minimum cost. They serve to illustrate how to adjust inputs to get the best results, like tuning a recipe for perfect cookies. When it comes to common mistakes, many people forget to consider the domain of \( f(x, y) \) which can lead to undefined values, especially with logarithmic functions. Just remember, the argument of a logarithm must always be positive!