Q.Find the derivative of $f(x) = x^2$ from first principles.

We apply the first-principles definition directly: form the difference quotient $\frac{f(x+h)-f(x)}{h}$ and take the limit as $h\to0$.

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

Step 1 — Write the difference quotient. By definition, $f'(x) = \displaystyle\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$, with $f(x)=x^2$.

Step 2 — Expand $f(x+h)$. $f(x+h) = (x+h)^2 = x^2+2xh+h^2$.

Step 3 — Form the difference. $f(x+h)-f(x) = (x^2+2xh+h^2)-x^2 = 2xh+h^2$.

Step 4 — Divide by $h$. $\dfrac{f(x+h)-f(x)}{h} = \dfrac{2xh+h^2}{h} = 2x+h$ (dividing every term by $h$, valid since $h\neq0$ in the limit process).

Step 5 — Let $h\to0$. $f'(x) = \displaystyle\lim_{h\to0}(2x+h) = 2x$.

Check (independent verification). This matches the general power rule $\dfrac{d}{dx}(x^n)=nx^{n-1}$ with $n=2$: $2x^{2-1}=2x$ — the first-principles working and the standard formula agree exactly.

A very common slip is dividing incorrectly in Step 4 — e.g. forgetting to divide the $h^2$ term by $h$, or cancelling $h$ from only part of the numerator. Another is substituting $h=0$ too early, before the $h$ in the denominator has been cancelled, which produces an undefined $\frac00$ instead of the correct limit.