Q.Differentiate $f(x) = x^3 - x$ from first principles.

Differentiating from first principles means applying the limit definition of the derivative directly, using the binomial expansion of $(x+h)^3$.

$f'(x) = \lim_{h\to0}\dfrac{\big[(x+h)^3-(x+h)\big]-\big[x^3-x\big]}{h} = 3x^2-1$

Given: $f(x) = x^3-x$; differentiate from first principles.

Step 1 — Write the limit definition: $f'(x) = \lim_{h\to0}\dfrac{f(x+h)-f(x)}{h} = \lim_{h\to0}\dfrac{\big[(x+h)^3-(x+h)\big]-\big[x^3-x\big]}{h}$.

Step 2 — Expand $(x+h)^3$: $(x+h)^3 = x^3+3x^2h+3xh^2+h^3$.

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

Step 4 — Subtract $f(x)=x^3-x$: $f(x+h)-f(x) = 3x^2h+3xh^2+h^3-h$.

Step 5 — Divide by $h$: $\dfrac{3x^2h+3xh^2+h^3-h}{h} = 3x^2+3xh+h^2-1$.

Step 6 — Let $h\to0$: every remaining term containing $h$ vanishes, leaving $f'(x) = 3x^2-1$.

Check (independent method — power and sum/difference rule): $\dfrac{d}{dx}(x^3) = 3x^2$ and $\dfrac{d}{dx}(-x) = -1$, so $f'(x) = 3x^2-1$ — matches the first-principles result exactly.

Forgetting to carry the $-x$ term through the expansion of $f(x+h)$ (writing $f(x+h)$ as only $(x+h)^3$ and dropping the $-(x+h)$ part) — every term of the original function must be substituted with $x+h$, not just its highest-power term.