Show that $\frac{d}{dx}\left(\frac{\ln x}{x^4}\right) = \frac{1}{x^5} - \frac{4\ln x}{x^5}$. Given that the curve $y = \mathrm{f}(x)$ passes through the point $\left(1, \frac{3}{4}\right)$ and is such that $\mathrm{f}'(x) = \frac{\ln x}{x^5}$, find $\mathrm{f}(x)$.

1. Apply the quotient rule to differentiate the expression

Let $y = \frac{\ln x}{x^4}$.
Using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$:
$\frac{dy}{dx} = \frac{x^4 \cdot \frac{d}{dx}(\ln x) - \ln x \cdot \frac{d}{dx}(x^4)}{(x^4)^2}$

2. Differentiate the terms and simplify to the required form

$\frac{dy}{dx} = \frac{x^4 \left(\frac{1}{x}\right) - (\ln x)(4x^3)}{x^8}$
$= \frac{x^3 - 4x^3 \ln x}{x^8}$
$= \frac{x^3(1 - 4\ln x)}{x^8}$
$= \frac{1 - 4\ln x}{x^5}$
$= \frac{1}{x^5} - \frac{4\ln x}{x^5}$ (Shown)

1. Rearrange the result from part (a) to make $\frac{\ln x}{x^5}$ the subject

From (a), we have:
$\frac{d}{dx}\left(\frac{\ln x}{x^4}\right) = \frac{1}{x^5} - \frac{4\ln x}{x^5}$
Rearranging this gives:
$\frac{4\ln x}{x^5} = \frac{1}{x^5} - \frac{d}{dx}\left(\frac{\ln x}{x^4}\right)$
$\frac{\ln x}{x^5} = \frac{1}{4}x^{-5} - \frac{1}{4}\frac{d}{dx}\left(\frac{\ln x}{x^4}\right)$

2. Integrate to find the general form of $\mathrm{f}(x)$

$\mathrm{f}(x) = \int \frac{\ln x}{x^5} \, dx$
$= \int \left( \frac{1}{4}x^{-5} - \frac{1}{4}\frac{d}{dx}\left(\frac{\ln x}{x^4}\right) \right) dx$
$= \frac{1}{4} \left( \frac{x^{-4}}{-4} \right) - \frac{1}{4}\left(\frac{\ln x}{x^4}\right) + C$
$= -\frac{1}{16x^4} - \frac{\ln x}{4x^4} + C$

3. Use the given point to find the constant of integration $C$

Given that the curve passes through $\left(1, \frac{3}{4}\right)$:
$\mathrm{f}(1) = -\frac{1}{16(1)^4} - \frac{\ln 1}{4(1)^4} + C = \frac{3}{4}$
Since $\ln 1 = 0$:
$-\frac{1}{16} + C = \frac{3}{4}$
$C = \frac{3}{4} + \frac{1}{16} = \frac{12}{16} + \frac{1}{16} = \frac{13}{16}$

4. State the final function $\mathrm{f}(x)$

$\mathrm{f}(x) = -\frac{1}{16x^4} - \frac{\ln x}{4x^4} + \frac{13}{16}$