Find $\dfrac{d}{dx}\left(\dfrac{\ln x}{4x}\right). Hence, determine the range of values of $x$ for which the curve $y=\dfrac{\ln x}{4x}$ is a decreasing function. Leave your answer in terms of $e$.

1. Apply the quotient rule to differentiate the expression

Let $u = \ln x$ and $v = 4x$.
$\dfrac{du}{dx} = \dfrac{1}{x}$ and $\dfrac{dv}{dx} = 4$.
$\dfrac{d}{dx}\left(\dfrac{\ln x}{4x}\right) = \dfrac{4x\left(\dfrac{1}{x}\right) - (\ln x)(4)}{(4x)^2}$

2. Simplify the resulting expression

$= \dfrac{4 - 4\ln x}{16x^2}$
$= \dfrac{4(1 - \ln x)}{16x^2}$
$= \dfrac{1 - \ln x}{4x^2}$

1. State the derivative from part (i) and set up the inequality for a decreasing function

From (i), $\frac{dy}{dx} = \frac{1 - \ln x}{4x^2}$.
For a decreasing function, $\frac{dy}{dx} < 0$.
$\frac{1 - \ln x}{4x^2} < 0$

2. Solve the inequality to find the range of $x$

Since $\ln x$ is only defined for $x > 0$, the denominator $4x^2$ is always positive.
Therefore, the numerator must be negative:
$1 - \ln x < 0$
$\ln x > 1$
$x > e^1$
$x > e$