AI Tutor
If the derivative of x1x^1 is 11, and the derivative of x2x^2 is 2x2x, you might already guess that x3x^3 gives 3x23x^2. But in mathematics, guessing a pattern isn't enough — we need an airtight proof.
The derivative measures the instantaneous rate of change of a function f(x)f(x) at any point, defined formally by the limit of the average slope as the horizontal step hh shrinks to zero: f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.

Differentiating the Cubic: f(x) = x³

To find the derivative of f(x)=x3f(x) = x^3, we substitute (x+h)(x+h) into the function and expand the expression (x+h)3(x+h)^3 algebraically.
Multiplying (x+h)(x+h)2(x+h)(x+h)^2 yields (x+h)(x2+2xh+h2)=x3+3x2h+3xh2+h3(x+h)(x^2 + 2xh + h^2) = x^3 + 3x^2h + 3xh^2 + h^3.
Subtracting f(x)=x3f(x) = x^3 leaves 3x2h+3xh2+h33x^2h + 3xh^2 + h^3. Dividing each term by hh gives 3x2+3xh+h23x^2 + 3xh + h^2, which smoothly evaluates to 3x23x^2 as h→0h \to 0.
Expanding (x+h)3(x+h)^3 was manageable by hand — but what if we need the derivative of x100x^{100} or xnx^n for any positive integer nn?

The Binomial Expansion Engine

To expand (x+h)n(x+h)^n, we use the Binomial Theorem, a algebraic formula established by Blaise Pascal and later generalized by Isaac Newton in 1665 to describe the expansion of powers of a sum.
The expansion states (x+h)n=xn+nxn−1h+(n2)xn−2h2+⋯+hn(x+h)^n = x^n + n x^{n-1}h + \binom{n}{2}x^{n-2}h^2 + \dots + h^n, where (nk)\binom{n}{k} represents the binomial coefficient, the number of ways to choose kk items from nn terms.
Notice a critical pattern: the first term has no hh, the second term has exactly h1h^1 with coefficient nxn−1n x^{n-1}, and every subsequent term contains h2h^2 or higher powers of hh.