Given $f(x)=\dfrac{2x-3}{x^2}$ , differentiate from first principles to find $f'(x)$ .

Model answer

(-2x+6)/x^3

Also accepted: -2x^-2+6x^-3, (6-2x)/x^3

Explanation: Using

f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}

and simplifying gives

f'(x)=\dfrac{-2x+6}{x^3}

, matching direct differentiation of

f(x)=2x^{-1}-3x^{-2}

Derived from ZIMSEC Pure Mathematics Paper 2, November 2025, Q3

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