Finding where x ln x is decreasing using the derivative

Key concept: This calculus question asks for the interval of decrease of a function defined on $x>0$. The solution depends on taking the derivative, analyzing its sign, and linking that sign to monotonicity.

Step by step

Differentiate the function

The function is

$f(x)=x\ln x,\quad x>0.$

Use the product rule:

$f'(x)=1\cdot \ln x + x\cdot \frac{1}{x}.$

So

$f'(x)=\ln x+1.$

Find where the derivative is negative

A function is decreasing where its derivative is negative. So we solve

$\ln x+1<0.$

This gives

$\ln x<-1,$

and exponentiating both sides yields

$x<e^{-1}=\frac{1}{e}.$

Because the domain is $x>0$, the interval of decrease is

$\boxed{0<x<\frac{1}{e}}.$

Match to the correct choice

The correct option is

$\boxed{A}.$

Why the sign of the derivative matters

For one-variable calculus, the derivative tells you the slope of the tangent line. When $f'(x)<0$, the graph slopes downward as $x$ increases, which means the function is decreasing. Here, $\ln x+1$ crosses zero at $x=1/e$, so the function decreases to the left of that point and increases to the right.

A quick sign test confirms this: if $x=0.1$, then $\ln(0.1)+1<0$; if $x=1$, then $\ln(1)+1=1>0$. That pattern matches the interval found above.

Pitfall alert

A common mistake is differentiating $x\ln x$ as $\ln x/x$ or forgetting the product rule entirely. Another issue is solving $\ln x+1<0$ and writing $x< -1/e$, which is impossible because $x>0$ on the domain. The exponential step must be handled carefully: from $\ln x<-1$, you get $x<e^{-1}$, not $x<-e^{-1}$. It is also easy to forget the domain restriction $x>0$, but that restriction is essential because $\ln x$ is not defined for nonpositive $x$.

Try different conditions

If the function were $f(x)=x\ln x-2x$, then the derivative would be $f'(x)=\ln x-1$, and the function would be decreasing when $\ln x<1$, that is, $0<x<e$. If the question asked for where the function is increasing instead, you would solve $\ln x+1>0$, giving $x>1/e$. The method stays the same: differentiate first, then analyze the sign of the derivative on the domain.

Further reading

product rule, derivative sign test, interval of decrease