## Calculus (3rd Edition)

f'(x)=$\ln$(x)+1
To differentiate the function y=x$\ln$(x) we will use the product rule which states that if f(x)=h(x)g(x), then f'(x)=g(x)h'(x)+h(x)g'(x). Therefore, we'll set h(x)=x and g(x)=$\ln$(x) and apply the product rule. Therefore, f'(x)=$\ln$(x)$\frac{d}{dx}$[x]+(x)$\frac{d}{dx}$[$\ln$(x)] $\frac{d}{dx}$[x]=1, using the power rule $\frac{d}{dx}$[$\ln$(x)]=$\frac{1}{x}$, according to the rule of the derivative of a natural log function f'(x)=$\ln$(x)+1