By differentiating from first principles, find the gradient function of the function f(x) = (2x−3)/x². Hence, or otherwise, find the equation of the tangent to the curve f(x) = (2x−3)/x² at the point where x = 2.

Given $f(x) = \frac{2x-3}{x^2}$, then $f(x+h) = \frac{2(x+h)-3}{(x+h)^2} = \frac{2x+2h-3}{(x+h)^2}$.

$$f'(x) = \lim_{h \to 0} \frac{\frac{2x+2h-3}{(x+h)^2} - \frac{2x-3}{x^2}}{h}$$

$$f'(x) = \lim_{h \to 0} \frac{1}{h} \left[ \frac{x^2(2x+2h-3) - (2x-3)(x+h)^2}{x^2(x+h)^2} \right]$$

Expand the terms in the numerator:

$$x^2(2x+2h-3) = 2x^3+2hx^2-3x^2$$ $$(2x-3)(x+h)^2 = (2x-3)(x^2+2xh+h^2) = 2x^3+4hx^2+2h^2x - 3x^2-6xh-3h^2$$

Subtract the expanded terms:

$$(2x^3+2hx^2-3x^2) - (2x^3+4hx^2+2h^2x - 3x^2-6xh-3h^2)$$ $$= 2x^3+2hx^2-3x^2 - 2x^3-4hx^2-2h^2x + 3x^2+6xh+3h^2$$ $$= -2hx^2 - 2h^2x + 6xh + 3h^2$$

$$f'(x) = \lim_{h \to 0} \frac{h(-2x^2 - 2hx + 6x + 3h)}{h \cdot x^2(x+h)^2}$$ $$f'(x) = \lim_{h \to 0} \frac{-2x^2 - 2hx + 6x + 3h}{x^2(x+h)^2}$$

$$f'(x) = \frac{-2x^2 - 2x(0) + 6x + 3(0)}{x^2(x+0)^2}$$ $$f'(x) = \frac{-2x^2 + 6x}{x^2(x^2)}$$ $$f'(x) = \frac{-2x^2 + 6x}{x^4}$$

Factor out $x$ from the numerator and simplify:

$$f'(x) = \frac{x(-2x + 6)}{x^4}$$ $$f'(x) = \boxed{\frac{6-2x}{x^3}}$$

Substitute $x=2$ into $f(x)$:

$$f(2) = \frac{2(2)-3}{2^2} = \frac{4-3}{4} = \frac{1}{4}$$

The point of tangency is $\left(2, \frac{1}{4}\right)$.

Use the gradient function $f'(x) = \frac{6-2x}{x^3}$ found in part (a).

$$m = f'(2) = \frac{6-2(2)}{2^3} = \frac{6-4}{8} = \frac{2}{8} = \frac{1}{4}$$

$$y - \frac{1}{4} = \frac{1}{4}(x - 2)$$

Multiply by 4 to clear the fractions:

$$4y - 1 = x - 2$$

Rearrange into the form $y = mx+c$:

$$4y = x - 1$$ $$y = \frac{1}{4}x - \frac{1}{4}$$

The equation of the tangent is $\boxed{y = \frac{1}{4}x - \frac{1}{4}}$.