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derivative of xln(x)

What is the derivative of xln(x)?

The derivative of x\ln(x) is \ln(x) + 1.

Proof. Let f(x) = x and g(x) = \ln(x). We will use the product rule:
\begin{align*}
(f(x)g(x))' = f'(x)g(x) + f(x)g'(x).
\end{align*}
This would mean that
\begin{align*}
f'(x) = 1,
\end{align*}
and
\begin{align*}
g'(x) = \frac{1}{x}
\end{align*}
as we have shown here. Wrapping everything together, we get
\begin{align*}
(f(x)g(x))' &= f'(x)g(x) + f(x)g'(x) \\
&= 1\cdot \ln(x) + x\frac{1}{x}\\
&= \ln(x) + 1.
\end{align*}
Therefore, we have that the derivative of x\ln(x) is \ln(x) + 1.

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