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Fundamental Theorem of Calculus & Key Derivatives

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1. Overview & Statement

The Fundamental Theorem of Calculus (FTC) bridges differential calculus and integral calculus, showing that differentiation and integration are inverse operations.

Part 1: Derivative of an Integral

If $f$ is continuous on $[a, b]$, then the function $g$ defined by:
$$g(x) = \int_{a}^{x} f(t) \, dt \quad \text{for } a \le x \le b$$
is continuous on $[a, b]$, differentiable on $(a, b)$, and its derivative is:
$$g'(x) = f(x)$$

Part 2: Evaluation Theorem

If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ ($F' = f$), then:
$$\int_{a}^{b} f(x) \, dx = F(b) - F(a)$$


2. High-Yield Derivative Rules Table

Function $f(x)$ Derivative $f'(x)$ Condition / Note
$x^n$ $n x^{n-1}$ Power Rule
$e^{kx}$ $k e^{kx}$ Exponential
$\ln(x)$ $\frac{1}{x}$ $x > 0$
$\sin(x)$ $\cos(x)$ Trigonometric
$\cos(x)$ $-\sin(x)$ Note the negative sign
$\tan(x)$ $\sec^2(x)$ Standard identity
$u(x) \cdot v(x)$ $u'v + uv'$ Product Rule
$\frac{u(x)}{v(x)}$ $\frac{u'v - uv'}{v^2}$ Quotient Rule ($v \ne 0$)

3. The Chain Rule

When dealing with composite functions $y = f(g(x))$:
$$\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$$

Quick Example:

For $y = \sin(3x^2 + 1)$:
1. Outer function: $\sin(u) \implies \cos(u)$
2. Inner function: $u = 3x^2 + 1 \implies \frac{du}{dx} = 6x$
3. Combined result:
$$\frac{dy}{dx} = 6x \cos(3x^2 + 1)$$


4. Key Exam Pitfalls to Avoid

  • Constant of Integration: Never forget $+ C$ for indefinite integrals.
  • Quotient Rule Ordering: Remember that $u'v - uv'$ has the numerator's derivative first.
  • Logarithmic Derivatives: $\frac{d}{dx} \ln(f(x)) = \frac{f'(x)}{f(x)}$ via chain rule.