Differentiate f(x) = x^3 - 3x^2 + 2x, and explain why x^3 becomes 3x^2.

Step-by-step solution

1. 1

   State the power rule and read what it does. For any constant exponent $n$,

   $\frac{d}{dx}\left(x^{n}\right)=n\,x^{n-1}$

   Two things happen at once: the exponent drops down in front as a multiplier, and the exponent decreases by one. That single sentence explains the $x^{3}\to 3x^{2}$ step people usually get stuck on.

2. 2

   Differentiate the cubic term. With $n=3$:

   $\frac{d}{dx}\left(x^{3}\right)=3x^{3-1}=3x^{2}$

   The $3$ is not invented — it is the old exponent, moved into the coefficient slot.

3. 3

   Differentiate the quadratic term, keeping the coefficient outside. Constants factor straight through a derivative, so

   $\frac{d}{dx}\left(-3x^{2}\right)=-3\cdot\frac{d}{dx}\left(x^{2}\right)=-3(2x)=-6x$

4. 4

   Differentiate the linear term. Here $n=1$, so $x^{1}\to 1\cdot x^{0}=1$ and

   $\frac{d}{dx}\left(2x\right)=2\cdot 1=2$

   A linear term always leaves behind just its slope.

5. 5

   Add the three pieces, using linearity of the derivative.

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

6. 6

   Check numerically at a point. A central difference at $x=2$ with $h=10^{-6}$ gives $\dfrac{f(2+h)-f(2-h)}{2h}\approx 2.000000$, and the formula gives $3(4)-6(2)+2=2$. The same check at $x=-1$ gives $\approx 11.000000$ against $3+6+2=11$.