The power rule, and where it stops AB & BC

For any real number $n$,

$$\frac{d}{dx}\left[x^{n}\right] = n\,x^{\,n-1}.$$

Bring the exponent down as a coefficient, then reduce the exponent by one. It is proved for positive integers directly from the difference quotient, and it holds far more generally.

Two special cases fall straight out and are worth memorizing as facts rather than rederiving each time:

1. $\frac{d}{dx}[x] = 1$, since $x = x^1$ and $1 \cdot x^0 = 1$.
2. $\frac{d}{dx}[c] = 0$ for any constant $c$. A constant function has a horizontal graph, so every tangent has slope zero.

The second one is not the power rule at all, and treating it as one produces $\frac{d}{dx}[5] = 5x^0 = 5$, which claims a constant function is rising.

The rule only recognizes something written as $x$ to a power. Most of the work in this topic is converting the expression into that shape before differentiating.

1. $\sqrt{x} = x^{1/2}$, so the derivative is $\frac{1}{2}x^{-1/2} = \dfrac{1}{2\sqrt{x}}$.
2. $\dfrac{1}{x^{3}} = x^{-3}$, so the derivative is $-3x^{-4} = \dfrac{-3}{x^{4}}$.
3. $\sqrt[3]{x^{2}} = x^{2/3}$, so the derivative is $\frac{2}{3}x^{-1/3}$.
4. $\dfrac{1}{\sqrt{x}} = x^{-1/2}$, so the derivative is $-\frac{1}{2}x^{-3/2}$.

Subtracting 1 from a negative exponent makes it more negative: $-3$ becomes $-4$, not $-2$. Subtracting 1 from a fraction needs a common denominator: $\frac{1}{2} - 1 = -\frac{1}{2}$ and $\frac{2}{3} - 1 = -\frac{1}{3}$.

Writing the answer back in radical or fraction form is optional unless the question asks for it, but the sign and the exponent have to be right either way.

The power rule requires a constant exponent on a variable base. Swap those roles and it says nothing.

$$\frac{d}{dx}\left[e^{x}\right] \neq x\,e^{\,x-1}, \qquad \frac{d}{dx}\left[2^{x}\right] \neq x\,2^{\,x-1}.$$

Those functions have their own derivatives, taken up in Topic 2.7: $\frac{d}{dx}[e^x] = e^x$ and $\frac{d}{dx}[a^x] = a^x \ln a$. The reason is that in $x^n$ the base moves as $x$ moves and the exponent is frozen, while in $a^x$ it is the other way around, which is a completely different kind of growth.

The check is quick: find the $x$. If it is in the base, the power rule applies. If it is in the exponent, it does not. If it is in both, as in $x^x$, neither elementary rule applies and a technique from Unit 3 is needed.

A constant multiplier is carried through untouched, which Topic 2.6 states as a rule in its own right. Applied to a power:

$$\frac{d}{dx}\left[7x^{3}\right] = 7 \cdot 3x^{2} = 21x^{2}.$$

The coefficient is multiplied by the exponent, never dropped and never differentiated on its own. Losing the 7 gives $3x^2$; differentiating it as though it were a second factor gives 0.

Three quick self-checks before moving on:

1. The exponent went down by exactly one. Including when it was negative or fractional.
2. The old exponent appears out front, multiplied into whatever coefficient was already there.
3. The base is still $x$. If the original had a variable exponent, none of the above applies.