Find the derivative \( f(x)=\ln \left(x e^{x}+e^{x^{2}}\right) \) a) \( y \prime=\frac{e^{x}(x+1)+2 x e^{x^{2}}}{x e^{x}+e^{x^{2}}} \) b) \( y \prime=\frac{2\left(e^{x}+x e^{x^{2}}\right)}{x e^{x}+e^{x^{2}}} \) c) \( y \prime=\frac{e^{x}+2 e^{x^{2}}}{x e^{x}+e^{x^{2}}} \) \( y \)

To find the derivative of \( f(x) = \ln \left(x e^{x} + e^{x^{2}}\right) \), we can use the chain rule along with the product and sum rules. First, let's differentiate the inside function using the quotient rule: \[ f'(x) = \frac{1}{x e^{x} + e^{x^{2}}} \cdot \left(\frac{d}{dx}(x e^{x}) + \frac{d}{dx}(e^{x^{2}})\right) \] Differentiating \( x e^{x} \) gives us \( e^{x} + x e^{x} \) (using the product rule), and differentiating \( e^{x^{2}} \) gives \( 2x e^{x^{2}} \) (using the chain rule). Thus, combining these results: \[ f'(x) = \frac{e^{x}(x + 1) + 2x e^{x^{2}}}{x e^{x} + e^{x^{2}}} \] So, the correct answer is (a) \( y' = \frac{e^{x}(x + 1) + 2x e^{x^{2}}}{x e^{x} + e^{x^{2}}} \). For some insights: Did you know that logarithmic functions, like the natural logarithm, have been historically significant? They were used in the 17th century for calculations, simplifying multiplications and divisions into additions and subtractions, making complex calculations easier! If you're looking to deepen your understanding of derivatives, exploring the rules of differentiation—such as product, quotient, and chain rules—is essential. A great resource for this is “Calculus Made Easy” by Silvanus P. Thompson, which presents complex ideas in a friendly manner!