Answer: \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{2(2x^3+5)}{x^3}\).

First rewrite the expression using powers of \(x\):

\(y=\dfrac{4x^3-5}{x^2}=4x-5x^{-2}.\)

Now differentiate term by term:

\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=4+10x^{-3}.\)

Write this with a common denominator of \(x^3\):

\(4+10x^{-3}=4+\dfrac{10}{x^3}=\dfrac{4x^3+10}{x^3}.\)

Factorise the numerator:

\(\dfrac{4x^3+10}{x^3}=\dfrac{2(2x^3+5)}{x^3}.\)

Therefore

\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{2(2x^3+5)}{x^3}.\)