Q.Find the derivative of $f(x) = \dfrac{1}{x}$ from first principles.

As before, we build the difference quotient for this rational function and simplify algebraically (combining the two fractions) before letting $h\to0$.

$f(x+h)-f(x) = \dfrac{1}{x+h}-\dfrac1x = \dfrac{x-(x+h)}{x(x+h)} = \dfrac{-h}{x(x+h)}$, so $\dfrac{f(x+h)-f(x)}{h} = \dfrac{-1}{x(x+h)}$. Letting $h\to0$: $f'(x) = -\dfrac{1}{x^2}$.

Step 1 — Write the difference quotient. $f'(x) = \displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$, with $f(x)=\dfrac1x$.

Step 2 — Combine the fractions in the numerator.

Step 3 — Divide by $h$.

(the $h$ in the numerator cancels with the $h$ we are dividing by).

Step 4 — Let $h\to0$. As $h\to0$, $x(x+h) \to x\cdot x = x^2$, so

Check (independent verification). Writing $f(x)=x^{-1}$ and applying the general power rule (quoted in the standard-derivatives section): $\dfrac{d}{dx}(x^{-1}) = -1\cdot x^{-2} = -\dfrac{1}{x^2}$ — matching the first-principles result exactly.

A common algebra slip is getting the sign wrong when combining $\frac{1}{x+h}-\frac1x$ into a single fraction — the numerator is $x-(x+h)=-h$, not $h$, and dropping that minus sign gives a wrong-signed final derivative. Students also sometimes forget to let $h\to0$ in the remaining $(x+h)$ term after cancelling, leaving an unsimplified expression instead of $x^2$.