Chapter 5: Basic Derivatives

We begin with the definition of the derivative:

$$\dfrac{d}{dx}x^n = \lim_{h \to 0} \dfrac{(x+h)^n - x^n}{h}$$

Step 1: Expand $(x+h)^n$ using the binomial theorem.

By the binomial theorem:

$$(x+h)^n = \displaystyle\sum_{k=0}^{n} \binom{n}{k} x^{n-k} h^k$$ $$= \binom{n}{0}x^n + \binom{n}{1}x^{n-1}h + \binom{n}{2}x^{n-2}h^2 + \cdots + \binom{n}{n}h^n$$ $$= x^n + nx^{n-1}h + \dfrac{n(n-1)}{2}x^{n-2}h^2 + \cdots + h^n$$

Step 2: Subtract $x^n$.

$$(x+h)^n - x^n = nx^{n-1}h + \dfrac{n(n-1)}{2}x^{n-2}h^2 + \cdots + h^n$$

Step 3: Divide by $h$.

$$\dfrac{(x+h)^n - x^n}{h} = nx^{n-1} + \dfrac{n(n-1)}{2}x^{n-2}h + \cdots + h^{n-1}$$

Step 4: Take the limit as $h \to 0$.

Every term containing $h$ approaches $0$, so:

$$\lim_{h \to 0} \dfrac{(x+h)^n - x^n}{h} = nx^{n-1}$$

By the definition of the derivative:

$$\dfrac{d}{dx}[c \cdot f(x)] = \lim_{h \to 0} \dfrac{c \cdot f(x+h) - c \cdot f(x)}{h}$$

Factor out $c$:

$$= \lim_{h \to 0} c \cdot \dfrac{f(x+h) - f(x)}{h}$$

A constant can be taken outside the limit:

$$= c \cdot \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$$

By the definition of $f'(x)$, this is:

$$= c \cdot f'(x)$$

By the definition of the derivative:

$$\dfrac{d}{dx}[f(x) + g(x)] = \lim_{h \to 0} \dfrac{[f(x+h) + g(x+h)] - [f(x) + g(x)]}{h}$$

Rearrange the terms:

$$= \lim_{h \to 0} \dfrac{[f(x+h) - f(x)] + [g(x+h) - g(x)]}{h}$$

Separate the sum:

$$= \lim_{h \to 0} \left[ \dfrac{f(x+h) - f(x)}{h} + \dfrac{g(x+h) - g(x)}{h} \right]$$

The limit distributes over addition:

$$= \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h} + \lim_{h \to 0} \dfrac{g(x+h) - g(x)}{h}$$

By the definition of the derivatives:

$$= f'(x) + g'(x)$$

Apply the sum rule and constant multiple rule repeatedly:

$$f'(x) = \dfrac{d}{dx}(a_n x^n) + \dfrac{d}{dx}(a_{n-1} x^{n-1}) + \cdots + \dfrac{d}{dx}(a_1 x) + \dfrac{d}{dx}(a_0)$$

Apply the constant multiple rule to each term:

$$= a_n \dfrac{d}{dx}(x^n) + a_{n-1} \dfrac{d}{dx}(x^{n-1}) + \cdots + a_1 \dfrac{d}{dx}(x) + 0$$

Apply the power rule:

$$= a_n \cdot nx^{n-1} + a_{n-1} \cdot (n-1)x^{n-2} + \cdots + a_1 \cdot 1$$

Simplifying gives:

$$= na_n x^{n-1} + (n-1)a_{n-1} x^{n-2} + \cdots + a_1$$

For $f(x) = x^{-n} = \dfrac{1}{x^n}$, the definition of the derivative gives:

$$f'(x) = \lim_{h \to 0} \dfrac{\dfrac{1}{(x+h)^n} - \dfrac{1}{x^n}}{h}$$

Step 1: Combine the fractions in the numerator.

$$= \lim_{h \to 0} \dfrac{1}{h} \cdot \dfrac{x^n - (x+h)^n}{(x+h)^n \cdot x^n}$$

Step 2: Expand $(x+h)^n$ in the numerator using the binomial theorem.

$$(x+h)^n = x^n + nx^{n-1}h + \text{(terms involving $h^2$ and higher powers)}$$

Therefore:

$$x^n - (x+h)^n = -nx^{n-1}h - \text{(terms involving $h^2$ and higher powers)}$$

Step 3: Substitute the expansion and divide by $h$.

$$f'(x) = \lim_{h \to 0} \dfrac{-nx^{n-1}h - (\text{terms involving $h^2$ and higher powers})}{h \cdot (x+h)^n \cdot x^n}$$ $$= \lim_{h \to 0} \dfrac{-nx^{n-1} - (\text{terms involving $h$ and higher powers})}{(x+h)^n \cdot x^n}$$

Step 4: Take the limit as $h \to 0$.

$$= \dfrac{-nx^{n-1}}{x^n \cdot x^n} = \dfrac{-nx^{n-1}}{x^{2n}} = -nx^{n-1-2n} = -nx^{-n-1}$$