Derivative Rules: Power, Product, Quotient, Chain

$$\frac{d}{dx}\left[x^n\right] = nx^{n-1} \qquad \frac{d}{dx}[c] = 0$$

Bring the exponent down in front, then subtract 1 from the exponent. The derivative of any constant is 0, because a constant never changes.

Example: Differentiate $x^5$, $\sqrt{x}$, and $\frac{1}{x^2}$.

1. $\frac{d}{dx}[x^5] = 5x^4$.
2. Rewrite $\sqrt{x}$ as $x^{1/2}$. Then the derivative is $\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.
3. Rewrite $\frac{1}{x^2}$ as $x^{-2}$. Then the derivative is $-2x^{-3} = -\frac{2}{x^3}$.

The power rule works for any real exponent, as long as you rewrite roots and fractions as powers first.

$$\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x) \qquad \frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$$

Constants in front just come along for the ride, and you can differentiate a sum (or difference) one term at a time.

Example: Differentiate $f(x) = 4x^3 - 5x^2 + 2x - 9$.

1. $4x^3$ becomes $4 \cdot 3x^2 = 12x^2$.
2. $-5x^2$ becomes $-5 \cdot 2x = -10x$.
3. $2x$ becomes $2$.
4. $-9$ is a constant, so it becomes $0$.
5. Put it together: $f'(x) = 12x^2 - 10x + 2$.

$$\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$$

Say it as "derivative of the first times the second, plus the first times derivative of the second."

Example: Differentiate $h(x) = (x^2 + 1)(x^3)$.

1. Let $f = x^2 + 1$ and $g = x^3$. Then $f' = 2x$ and $g' = 3x^2$.
2. Apply the rule: $h'(x) = 2x \cdot x^3 + (x^2 + 1) \cdot 3x^2$.
3. Simplify: $2x^4 + 3x^4 + 3x^2 = 5x^4 + 3x^2$.

Check: expanding first gives $h(x) = x^5 + x^3$, and its derivative is $5x^4 + 3x^2$. Same answer.

Another one: $\frac{d}{dx}[x^2 \sin x] = 2x \sin x + x^2 \cos x$.

$$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$$

A common memory trick is "low d-high minus high d-low, over low squared." The order in the numerator matters because of the minus sign.

Example: Differentiate $h(x) = \frac{x^2}{x + 1}$.

1. Top: $f = x^2$, so $f' = 2x$. Bottom: $g = x + 1$, so $g' = 1$.
2. Apply the rule: $h'(x) = \frac{2x(x + 1) - x^2(1)}{(x + 1)^2}$.
3. Expand the numerator: $2x^2 + 2x - x^2 = x^2 + 2x$.
4. Final answer: $h'(x) = \frac{x^2 + 2x}{(x + 1)^2}$.

$$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$$

Use the chain rule for a function inside another function. Differentiate the outside, leave the inside alone, then multiply by the derivative of the inside.

Example 1: Differentiate $(3x^2 + 1)^4$.

1. Outside: $u^4$, with inside $u = 3x^2 + 1$.
2. Derivative of the outside: $4(3x^2 + 1)^3$.
3. Derivative of the inside: $6x$.
4. Multiply: $4(3x^2 + 1)^3 \cdot 6x = 24x(3x^2 + 1)^3$.

Example 2: Differentiate $\sqrt{x^2 + 9}$.

1. Rewrite as $(x^2 + 9)^{1/2}$.
2. Outside derivative: $\frac{1}{2}(x^2 + 9)^{-1/2}$.
3. Inside derivative: $2x$.
4. Multiply and simplify: $\frac{2x}{2\sqrt{x^2 + 9}} = \frac{x}{\sqrt{x^2 + 9}}$.

Real problems often need more than one rule. Work from the outside structure in.

Example: Differentiate $x^3 e^{2x}$.

1. It is a product, so use the product rule with $f = x^3$ and $g = e^{2x}$.
2. $f' = 3x^2$. For $g'$, use the chain rule: $e^{2x} \cdot 2 = 2e^{2x}$.
3. Product rule: $3x^2 e^{2x} + x^3 \cdot 2e^{2x}$.
4. Factor out $x^2 e^{2x}$: the answer is $x^2 e^{2x}(3 + 2x)$.

Paste it into the VeilGPT solver to check your steps on harder combinations.