If $h(x)=x \ln x$ , find $h'(x)$.

Question

I'm not sure how to differentiate this function. I guess you use the chain rule, but I am not getting the correct answer. Could someone show me how it's done?

Answer 1

Use the product rule $(uv)' = u'v + uv'$:

$$\left ( x \ln x \right)' = x' \ln x + x (\ln x)' = \ln x + {x \over x} = \ln x + 1 $$

Answer 2

Simply use product rule: $$h(x)=x\ln x$$ $$h'(x)=x\ln' x+x'\ln x$$ $\ln'(x)=\frac 1 x$ and $x'=1$, so substitute: $$h'(x)=\frac x x+1\ln x$$ Simplify: $$h'(x)=1+\ln x$$