CE Board Exam Randomizer

⬅ Back to Subject Topics

Differential Calculus

Limits and Continuity

Limits describe the behavior of a function as the input approaches a value. Continuity means the limit equals the function value at that point.

$$\lim_{x \to a} f(x) = L$$
$$\text{Continuous at } x=a \iff \begin{cases} f(a) \text{ is defined} \\ \lim_{x \to a} f(x) \text{ exists} \\ \lim_{x \to a} f(x) = f(a) \end{cases} $$

Types of discontinuities: removable, jump, infinite.

Definition of the Derivative

The derivative measures instantaneous rate of change and slope of the tangent line.

$$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$
$$f'(a) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a}$$

Tangent and Normal Lines

1. Equation of the tangent line at $x=a$:

$$y - f(a) = f'(a)(x - a)$$

2. Equation of the normal line at $x=a$:

$$y - f(a) = -\frac{1}{f'(a)}(x - a)$$

Linear Approximation and Differentials — Key Formulas

$$L(x) = f(a) + f'(a)(x-a)$$
$$dy = f'(x)\, dx$$
$$\Delta y \approx dy$$

Basic Differentiation Rules

$$\frac{d}{dx}[c] = 0$$
$$\frac{d}{dx}[x^n] = nx^{n-1}$$
$$\frac{d}{dx}[c f(x)] = c f'(x)$$
$$\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)$$

Product, Quotient, and Chain Rules

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

Derivatives of Trigonometric Functions

$$\frac{d}{dx}(\sin x) = \cos x$$
$$\frac{d}{dx}(\cos x) = -\sin x$$
$$\frac{d}{dx}(\tan x) = \sec^2 x$$
$$\frac{d}{dx}(\cot x) = -\csc^2 x$$
$$\frac{d}{dx}(\sec x) = \sec x \tan x$$
$$\frac{d}{dx}(\csc x) = -\csc x \cot x$$

Derivatives of Inverse Trigonometric Functions

$$\frac{d}{dx}(\sin^{-1} u) = \frac{u'}{\sqrt{1 - u^2}}$$
$$\frac{d}{dx}(\cos^{-1} u) = \frac{-u'}{\sqrt{1 - u^2}}$$
$$\frac{d}{dx}(\tan^{-1} u) = \frac{u'}{1 + u^2}$$
$$\frac{d}{dx}(\cot^{-1} u) = \frac{-u'}{1 + u^2}$$
$$\frac{d}{dx}(\sec^{-1} u) = \frac{u'}{|u|\sqrt{u^2 - 1}}$$
$$\frac{d}{dx}(\csc^{-1} u) = \frac{-u'}{|u|\sqrt{u^2 - 1}}$$

Derivatives of Exponential and Logarithmic Functions

$$\frac{d}{dx}[e^x] = e^x$$
$$\frac{d}{dx}[\ln x] = \frac{1}{x}$$
$$\frac{d}{dx}[a^x] = a^x \ln a$$
$$\frac{d}{dx}[\log_a x] = \frac{1}{x \ln a}$$

Hyperbolic Functions

Certain combinations of the exponential functions $e^x$ and $e^{-x}$ occur frequently in mathematics, science, and engineering. These are called hyperbolic functions.

Definitions of hyperbolic functions:

$$ \sinh x = \frac{e^{x} - e^{-x}}{2} $$
$$ \cosh x = \frac{e^{x} + e^{-x}}{2} $$
$$ \tanh x = \frac{\sinh x}{\cosh x} $$
$$ coth x = \frac{\cosh x}{\sinh x} $$
$$ sech x = \frac{1}{\cosh x} $$
$$ csch x = \frac{1}{\sinh x} $$

Names are read as “hyperbolic sine,” “hyperbolic cosine,” and so on.

Fundamental Hyperbolic Identities

$$ \cosh^{2} x - \sinh^{2} x = 1 $$
$$ \tanh^{2} x + sech^{2} x = 1 $$
$$ coth^{2} x - csch^{2} x = 1 $$
$$ \sinh 2x = 2 \sinh x \cosh x $$
$$ \cosh 2x = \cosh^{2} x + \sinh^{2} x = 1 + 2\sinh^{2} x = 2\cosh^{2} x - 1 $$

Differentiation of Hyperbolic Functions

The derivative rules for hyperbolic functions (where $u$ is a function of $x$) are:

$$ \frac{d}{dx}(\sinh u) = \cosh u \, \frac{du}{dx} $$
$$ \frac{d}{dx}(\cosh u) = \sinh u \, \frac{du}{dx} $$
$$ \frac{d}{dx}(\tanh u) = sech^{2} u \, \frac{du}{dx} $$
$$ \frac{d}{dx}(coth u) = -csch^{2} u \, \frac{du}{dx} $$
$$ \frac{d}{dx}(sech u) = -sech u \tanh u \, \frac{du}{dx} $$
$$ \frac{d}{dx}(csch u) = -csch u \coth u \, \frac{du}{dx} $$

Higher Order and Implicit Differentiation

$$y' = \frac{dy}{dx}$$
$$y'' = \frac{d^2 y}{dx^2}$$
$$y^{(n)} = \frac{d^n y}{dx^n}$$

Implicit differentiation:

$$\frac{d}{dx}F(x,y) = F_x + F_y \frac{dy}{dx} = 0$$
$$\frac{dy}{dx} = -\frac{F_x}{F_y}$$

Linear Approximation and Differentials

$$dy = f'(x)\, dx$$
$$\Delta y \approx dy = f'(x)\Delta x$$
$$L(x) = f(a) + f'(a)(x-a)$$

Monotonicity and Extrema

$$f'(x) > 0 \Rightarrow \text{increasing}$$
$$f'(x) < 0 \Rightarrow \text{decreasing}$$

Concavity, Inflection Points, and the Second Derivative Test

The second derivative $f''(x)$ tells us how the graph of a function bends. Understanding concavity helps us identify inflection points and determine whether critical points are maxima or minima.

Concavity of the graph:

$$f''(x) > 0 \Rightarrow \text{the graph is concave up (shaped like a cup)}$$
$$f''(x) < 0 \Rightarrow \text{the graph is concave down (shaped like a hill)}$$

Inflection points:

An inflection point occurs where the graph changes concavity — from concave up to concave down or vice versa. This usually happens where $f''(x)$ changes sign.

$$\text{If } f''(x) \text{ changes sign at } x = c,\; \text{then } c \text{ is an inflection point.}$$

The Second Derivative Test for maxima and minima:

If a point $x = c$ is a critical point (meaning $f'(c) = 0$), then the second derivative helps determine whether it is a peak or a valley.

$$f''(c) > 0 \Rightarrow \text{local minimum at } c \quad (\text{graph curves upward})$$
$$f''(c) < 0 \Rightarrow \text{local maximum at } c \quad (\text{graph curves downward})$$
$$f''(c) = 0 \Rightarrow \text{test is inconclusive; use another method}$$

Taken together, concavity and the second derivative test provide a complete picture of how the function behaves: where it bends, where it reaches peaks and valleys, and where the direction of bending changes.

Optimization Problems (Maxima-Minima)

How to solve optimization problems:

  1. Write an equation for the quantity you want to maximize or minimize.
  2. Differentiate the equation and set $f'(x)=0$ to find possible maximum or minimum values.
  3. Check which value actually gives the largest or smallest result by comparing the outputs of $f(x)$.

These problems usually involve finding the best (maximum or minimum) value under certain conditions, such as fixed perimeter, fixed area or volume, limited materials, or being restricted to a given path or shape.

Steps in Solving Implicit Differentiation Problems

How to differentiate equations where $y$ is not isolated:

  1. Differentiate both sides of the equation with respect to $x$.
  2. Apply the chain rule whenever you differentiate a term involving $y$, which produces $\frac{dy}{dx}$.
  3. Collect all terms containing $\frac{dy}{dx}$ on one side of the equation.
  4. Factor out $\frac{dy}{dx}$.
  5. Solve for $\frac{dy}{dx}$.

Implicit differentiation is used when the equation cannot easily be solved for $y$, such as circles, ellipses, or equations where $x$ and $y$ are mixed together.

Radius of Curvature

The radius of curvature measures how sharply a curve bends at a point.

1. If the curve is given as $y = f(x)$:

$$R = \frac{\left(1 + (y')^2\right)^{3/2}}{|y''|}$$

2. If the curve is given parametrically as $x(t)$ and $y(t)$:

$$R = \frac{\left(\dot{x}^2 + \dot{y}^2\right)^{3/2}} {|\dot{x}\ddot{y} - \dot{y}\ddot{x}|}$$

3. If the curve is given in the form $x = g(y)$:

$$R = \frac{\left(1 + (x')^2\right)^{3/2}}{|x''|}$$

4. Curvature $\kappa$ (reciprocal of the radius of curvature):

$$\kappa = \frac{1}{R}$$

L'Hospital's Rule

$$\text{If } \lim_{x\to a} f(x)=0 \text{ and } \lim_{x\to a} g(x)=0$$
$$\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}$$

Mean Value Theorem

$$f \text{ continuous on } [a,b],\; f \text{ differentiable on } (a,b)$$
$$\exists c\in(a,b) \text{ such that } f'(c) = \frac{f(b)-f(a)}{b-a}$$

Rolle's Theorem: if f(a) = f(b), then there exists a number c with $f'(c)=0$.

Steps in Finding Partial Derivatives

How to compute partial derivatives of a function of two or more variables:

  1. Identify which variable you are differentiating with respect to (usually $x$, $y$, or $z$).
  2. Treat all other variables as constants. Only the chosen variable is allowed to change.
  3. Differentiate the function using ordinary differentiation rules (power rule, product rule, chain rule, etc.).
  4. Write the result using partial derivative notation such as $\frac{\partial f}{\partial x}$ or $f_x$.
  5. If needed, repeat the process to find higher-order partial derivatives like $f_{xx}$, $f_{yy}$, or mixed partials such as $f_{xy}$ and $f_{yx}$.

Notation for partial derivatives:

$$f_x = \frac{\partial f}{\partial x}$$
$$f_y = \frac{\partial f}{\partial y}$$
$$f_{xx} = \frac{\partial^2 f}{\partial x^2}$$
$$f_{xy} = \frac{\partial^2 f}{\partial x \partial y}$$

Partial derivatives appear in problems involving surfaces, optimization of functions of two variables, and models where a quantity depends on more than one input.

Concept Concept Concept Concept Concept Concept Concept Concept Concept

Problem:

Refer to the image shown:

Diagram Diagram Diagram

See images:

Solution Solution Solution Solution

Problem:

Refer to the image shown:

Diagram Diagram Diagram

See images:

Solution Solution Solution Solution

Problem:

Refer to the image shown:

Diagram Diagram Diagram

See images:

Solution Solution Solution Solution

Problem:

Refer to the image shown:

Diagram Diagram Diagram

See images:

Solution Solution Solution Solution

Problem:

Refer to the image shown:

Diagram Diagram Diagram

See images:

Solution Solution Solution Solution

Problem:

Refer to the image shown:

Diagram Diagram Diagram

See images:

Solution Solution Solution Solution

Problem:

Refer to the image shown:

Diagram Diagram Diagram

See images:

Solution Solution Solution Solution

Problem:

Refer to the image shown:

Diagram Diagram Diagram

See images:

Solution Solution Solution Solution
Scroll to zoom