Derivatives, Differentiation Rules, and Function Analysis
Modules 2, 3, & 4 - Complete Reference
This guide contains every rule, formula, and technique you need for calculus derivatives. Each section builds upon the previous one, so master the fundamentals before moving to advanced topics.
The derivative of a function $f(x)$ at point $x = a$ is:
$$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$Alternatively:
$$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$$The derivative represents the instantaneous rate of change or the slope of the tangent line to the curve at a given point.
Memory tip: The slope of a horizontal line is zero.
Process: Bring down the power, subtract 1 from the exponent.
Rule: Constants "factor out" of derivatives.
Rule: Derivative of a sum equals sum of derivatives.
Find $\frac{d}{dx}[(x^2 + 1)(3x - 2)]$
Solution:
$f(x) = x^2 + 1$, so $f'(x) = 2x$
$g(x) = 3x - 2$, so $g'(x) = 3$
Result: $2x(3x - 2) + (x^2 + 1)(3) = 6x^2 - 4x + 3x^2 + 3 = 9x^2 - 4x + 3$
Find $\frac{d}{dx}\left[\frac{x^2}{x+1}\right]$
Solution:
$f(x) = x^2$, so $f'(x) = 2x$
$g(x) = x+1$, so $g'(x) = 1$
Result: $\frac{2x(x+1) - x^2(1)}{(x+1)^2} = \frac{2x^2 + 2x - x^2}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2}$
If $h(x) = f(g(x))$, then:
$$h'(x) = f'(g(x)) \cdot g'(x)$$Alternative notation: If $y = f(u)$ and $u = g(x)$, then:
$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$Example 1: $\frac{d}{dx}[(x^2 + 3x)^5]$
Outer: $u^5$, Inner: $x^2 + 3x$
Result: $5(x^2 + 3x)^4 \cdot (2x + 3)$
Example 2: $\frac{d}{dx}[\sin(3x^2)]$
Outer: $\sin(u)$, Inner: $3x^2$
Result: $\cos(3x^2) \cdot 6x$
Use when the equation cannot be easily solved for $y$ in terms of $x$, or when both variables are mixed together.
Step 1: Differentiate both sides
$\frac{d}{dx}[x^2 + y^2] = \frac{d}{dx}[25]$
Step 2: Apply differentiation rules
$2x + 2y\frac{dy}{dx} = 0$
Step 3: Solve for $\frac{dy}{dx}$
$2y\frac{dy}{dx} = -2x$
$\frac{dy}{dx} = -\frac{x}{y}$
Differentiate: $3x^2 + 2xy + x^2\frac{dy}{dx} + 8y\frac{dy}{dx} = 0$
Factor: $3x^2 + 2xy + \frac{dy}{dx}(x^2 + 8y) = 0$
Solve: $\frac{dy}{dx} = -\frac{3x^2 + 2xy}{x^2 + 8y}$
Problems where you know the rate of change of one quantity and need to find the rate of change of a related quantity.
Problem: A spherical balloon is being inflated at 10 cm³/s. When radius = 8 cm, find the rate at which radius is increasing.
Given: $\frac{dV}{dt} = 10$ cm³/s, find $\frac{dr}{dt}$ when $r = 8$ cm
Equation: $V = \frac{4}{3}\pi r^3$
Differentiate: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$
Substitute: $10 = 4\pi(8)^2 \frac{dr}{dt} = 256\pi \frac{dr}{dt}$
Solve: $\frac{dr}{dt} = \frac{10}{256\pi} = \frac{5}{128\pi}$ cm/s
Problem: An 18-foot ladder leans against a wall. The base moves away from the wall at 5 ft/s. When the base is 4 feet from the wall, how fast is the top sliding down?
Setup: Let $x$ = distance from wall to base, $y$ = height of top
Given: $\frac{dx}{dt} = 5$ ft/s, find $\frac{dy}{dt}$ when $x = 4$
Equation: $x^2 + y^2 = 18^2 = 324$
When $x = 4$: $16 + y^2 = 324$, so $y = \sqrt{308} = 2\sqrt{77}$
Differentiate: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$
Substitute: $2(4)(5) + 2(2\sqrt{77})\frac{dy}{dt} = 0$
Solve: $\frac{dy}{dt} = -\frac{40}{4\sqrt{77}} = -\frac{10}{\sqrt{77}}$ ft/s
The linear approximation of $f(x)$ near $x = a$ is:
$$L(x) = f(a) + f'(a)(x - a)$$This gives the equation of the tangent line at $(a, f(a))$.
Use when you need to estimate function values near a point where you know both $f(a)$ and $f'(a)$, especially when exact computation is difficult.
Setup: Use $f(x) = \sqrt{x}$ near $a = 8$
$f(8) = \sqrt{8} = 2\sqrt{2} \approx 2.828$
$f'(x) = \frac{1}{2\sqrt{x}}$, so $f'(8) = \frac{1}{2\sqrt{8}} = \frac{1}{4\sqrt{2}}$
Linear approximation:
$L(x) = 2\sqrt{2} + \frac{1}{4\sqrt{2}}(x - 8)$
Estimate: $L(7.95) = 2\sqrt{2} + \frac{1}{4\sqrt{2}}(-0.05) \approx 2.819$
For small changes, we write:
$$dy = f'(x) dx$$This approximates the change in $y$ when $x$ changes by $dx$.
Point: $(2, f(2)) = (2, 8 - 8) = (2, 0)$
Slope: $f'(x) = 3x^2 - 4$, so $f'(2) = 12 - 4 = 8$
Equation: $y - 0 = 8(x - 2)$, so $y = 8x - 16$
For $f(x) = b^x$ where $b > 0, b \neq 1$:
$f(x) = e^x$ where $e \approx 2.71828...$
Key Property: $\frac{d}{dx}[e^x] = e^x$
The function $e^x$ is its own derivative!
Example 1: $\frac{d}{dx}[e^{3x^2}] = e^{3x^2} \cdot 6x$
Example 2: $\frac{d}{dx}[2^x] = 2^x \ln 2$
Example 3: $\frac{d}{dx}[e^{\sin x}] = e^{\sin x} \cdot \cos x$
$y = \log_a x$ if and only if $a^y = x$
Natural Logarithm: $\ln x = \log_e x$
Key Relationships:
For $x, y > 0$:
$$\log_a(xy) = \log_a x + \log_a y$$ $$\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y$$ $$\log_a(x^r) = r\log_a x$$Problem: Find $\frac{dy}{dx}$ if $y = x^x$
Step 1: Take natural log of both sides
$\ln y = \ln(x^x) = x \ln x$
Step 2: Differentiate both sides
$\frac{1}{y} \frac{dy}{dx} = \ln x + x \cdot \frac{1}{x} = \ln x + 1$
Step 3: Solve for $\frac{dy}{dx}$
$\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)$
If $f$ is continuous and one-to-one on an interval, and differentiable at $a$ with $f'(a) \neq 0$, then $f^{-1}$ is differentiable at $b = f(a)$ and:
$$(f^{-1})'(b) = \frac{1}{f'(f^{-1}(b))} = \frac{1}{f'(a)}$$Step 1: $y = \sqrt{x-2}$
Step 2: $x = \sqrt{y-2}$
Step 3: $x^2 = y-2$, so $y = x^2 + 2$
Step 4: $f^{-1}(x) = x^2 + 2$ for $x \geq 0$
Example 1: $\frac{d}{dx}[\sin^{-1}(3x)] = \frac{3}{\sqrt{1-(3x)^2}} = \frac{3}{\sqrt{1-9x^2}}$
Example 2: $\frac{d}{dx}[\tan^{-1}(x^2)] = \frac{2x}{1+(x^2)^2} = \frac{2x}{1+x^4}$
Formula:
$$L(x) = f(a) + f'(a)(x-a)$$Use when estimating function values near known points.