Comprehensive Calculus Study Guide: Differentiation Rules, Chain Rule, Implicit Differentiation, and Related Rates

Fundamental Differentiation Rules

• The Constant Rule

  • Theorem Statement: Let $c$ be any real constant. Then the derivative of $c$ with respect to $x$ is zero:     $\frac{d}{dx}(c) = 0$

  • Geometric Interpretation: The graph of a constant function $y = c$ is a horizontal straight line. Because the slope of any horizontal line is identically $0$ everywhere, the instantaneous rate of change (derivative) is always $0$.     

    

  • Worked Examples:

  • For $f(x) = 7$: $f'(x) = \frac{d}{dx}(7) = 0$.

  • For $y = \pi$: $\frac{dy}{dx} = \frac{d}{dx}(\pi) = 0$.

  • For $g(t) = e^2$: $g'(t) = \frac{d}{dt}(e^2) = 0$.

  • For $v = \cos(1)$: $\frac{dv}{ds} = \frac{d}{ds}(\cos(1)) = 0$.

  • For $f(x) = \sin\left(\frac{\pi}{2}\right)$: $\frac{df}{dx} = \frac{d}{dx}\left(\sin\left(\frac{\pi}{2}\right)\right) = 0$.

  • For $y = e^\pi - 5^2 + 2$: Since every term is a constant real number, $y' = 0$.

  • For $f(x) = \cos(3) - 2 + \pi^2 + \ln(7)$: All constituent terms are constants, so $f'(x) = 0 - 0 + 0 + 0 = 0$.

• The Power Rule

  • Theorem Statement: Let $n$ be any real number. Then the derivative of the power function $x^n$ with respect to $x$ is given by:     $\frac{d}{dx}(x^n) = n x^{n-1}$

  • Worked Examples:

  • Differentiating a radical term $\frac{d}{dx}\left(\sqrt{x^3}\right)$: Rewrite as a rational exponent $x^{3/2}$, yielding $\frac{d}{dx}\left(x^{3/2}\right) = \frac{3}{2} x^{(3/2) - 1} = \frac{3}{2} x^{1/2} = \frac{3}{2}\sqrt{x}$.

  • For $f(x) = x^{100}$: $f'(x) = 100 x^{99}$.

  • For $y = \frac{1}{x} = x^{-1}$: $\frac{dy}{dx} = (-1)x^{-1-1} = -x^{-2} = -\frac{1}{x^2}$.

  • For $s(t) = \frac{1}{t^3} = t^{-3}$: $\frac{ds}{dt} = -3 t^{-3-1} = -3 t^{-4} = -\frac{3}{t^4}$.

  • For $g(x) = \sqrt[3]{x^{50}} = x^{50/3}$: $g'(x) = \frac{50}{3} x^{(50/3) - 1} = \frac{50}{3} x^{47/3}$.

  • For $y = x^\pi$: $\frac{dy}{dx} = \pi x^{\pi - 1}$.

• The Constant Multiple Rule

  • Theorem Statement: Let $c$ be a constant and $f$ be a differentiable function. Then:     $\frac{d}{dx}(c f(x)) = c f'(x)$

  • Application:

  • For $y = 2x^4$: $\frac{dy}{dx} = 2 \cdot \frac{d}{dx}(x^4) = 2(4x^3) = 8x^3$.

• The Sum and Difference Rule

  • Theorem Statement: Let $f$ and $g$ be differentiable functions. Then:     $\frac{d}{dx}(f(x) \pm g(x)) = f'(x) \pm g'(x)$

  • Worked Examples with Algebraic Manipulation:

  • For $s(r) = r^5 - 3r^3 + r^2 - 7$: $\frac{ds}{dr} = 5r^4 - 9r^2 + 2r$.

  • For $y = x^2(x - x^3)$: Expand algebraically first to get $y = x^3 - x^5$, then differentiate to obtain $y' = 3x^2 - 5x^4$.

  • For $s(t) = t^{3/5} + t^{-2/7}$: $\frac{ds}{dt} = \frac{3}{5}t^{-2/5} - \frac{2}{7}t^{-9/7}$.

  • For $f(y) = \frac{y + 1}{y^2}$: Rewrite by splitting the fraction into $y^{-1} + y^{-2}$, yielding $f'(y) = -y^{-2} - 2y^{-3}$.

  • For $f(x) = \frac{2x^4 + \sqrt{x}}{3x}$: Split into individual terms and simplify exponents:       $f(x) = \frac{2x^4}{3x} + \frac{x^{1/2}}{3x} = \frac{2}{3}x^3 + \frac{1}{3}x^{-1/2}$       $f'(x) = \frac{2}{3}(3x^2) + \frac{1}{3}\left(-\frac{1}{2}\right)x^{-3/2} = 2x^2 - \frac{1}{6}x^{-3/2}$

  • For $r(\theta) = e^\theta + \theta^{-2} + \sqrt[4]{\theta} = e^\theta + \theta^{-2} + \theta^{1/4}$: $r'(\theta) = e^\theta - 2\theta^{-3} + \frac{1}{4}\theta^{-3/4}$.

• Exponential Derivatives

  • Theorem Statement: Let $b$ be a positive real number ($b > 0$). Then:     $\frac{d}{dx}(b^x) = b^x \ln(b)$

  • Natural Exponential Base ($b = e$): Because $\ln(e) = 1$, the derivative simplifies to:     $\frac{d}{dx}(e^x) = e^x$

  • Worked Examples:

  • For $y = \frac{2^x + 3^x}{5} = \frac{1}{5}(2^x) + \frac{1}{5}(3^x)$: $\frac{dy}{dx} = \frac{1}{5}\ln(2) 2^x + \frac{1}{5}\ln(3) 3^x$.

  • For $g(t) = \frac{e^{2t} - e^t}{3e^t}$: Simplify using exponent rules:       $g(t) = \frac{e^{2t}}{3e^t} - \frac{e^t}{3e^t} = \frac{1}{3}e^{2t - t} - \frac{1}{3}e^{t - t} = \frac{1}{3}e^t - \frac{1}{3}$       $g'(t) = \frac{1}{3}e^t - 0 = \frac{1}{3}e^t$

• Physical Kinematic Application: Position, Velocity, and Acceleration

  • Definitions: Given position $s(t)$, instantaneous velocity is $v(t) = s'(t) = \frac{ds}{dt}$, and instantaneous acceleration is $a(t) = v'(t) = s''(t) = \frac{dv}{dt}$.

  • Full Derivation Example: Let position be $s(t) = 3\sqrt[5]{t^8} + 5^t = 3t^{8/5} + 5^t$.

  • Instantaneous Velocity:       $v(t) = s'(t) = 3 \cdot \frac{d}{dt}\left(t^{8/5}\right) + \frac{d}{dt}\left(5^t\right)$       $v(t) = 3\left(\frac{8}{5}\right)t^{3/5} + 5^t \ln(5) = \frac{24}{5}t^{3/5} + 5^t \ln(5)$

  • Instantaneous Acceleration:       $a(t) = v'(t) = \frac{24}{5} \cdot \frac{d}{dt}\left(t^{3/5}\right) + \ln(5) \cdot \frac{d}{dt}\left(5^t\right)$       $a(t) = \frac{24}{5}\left(\frac{3}{5}\right)t^{-2/5} + \ln(5)\left(5^t \ln(5)\right) = \frac{72}{25}t^{-2/5} + 5^t \ln^2(5)$

The Chain Rule

• Theorem Formulation

  • Lagrange (Prime) Notation: If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$, then the composite function $F = f \circ g$, defined by $F(x) = f(g(x))$, is differentiable at $x$, and its derivative is:     $F'(x) = f'(g(x)) \cdot g'(x)$

  • Leibniz Notation: If $y = f(u)$ and $u = g(x)$ are both differentiable functions, then:     $\frac{dy}{dx} = \frac{df}{du} \cdot \frac{du}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

  • Intuition via Differentials: Treating Leibniz derivatives as formal ratios reveals the visual cancellation of intermediate differential terms:     $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

  • Multi-Layer Chain Rule Extension: When composing three or more functions, such as $y = f(g(h(x)))$, or across intermediate variables $u, v, w$:     $\frac{d}{dx}[f(g(h(x)))] = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$     $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dw} \cdot \frac{dw}{dx}$

• Algebraic and Transcendental Worked Examples

  • Polynomial Composite Power: For $h(x) = (x^2 + 2x)^3$:

  • Outer function: $f(u) = u^3$ with $u = g(x) = x^2 + 2x$.

  • Outer derivative: $f'(u) = 3u^2 = 3(x^2 + 2x)^2$.

  • Inner derivative: $g'(x) = 2x + 2$.

  • Final derivative: $h'(x) = 3(x^2 + 2x)^2 \cdot (2x + 2)$.

  • Trigonometric Power: For $y = \sin^2(x) = (\sin(x))^2$:

  • Inner function: $u = \sin(x)$; outer function: $y = u^2$.

  • Component derivatives: $\frac{dy}{du} = 2u$, $\frac{du}{dx} = \cos(x)$.

  • Combined derivative: $\frac{dy}{dx} = 2u \cos(x) = 2\sin(x)\cos(x)$.

  • Trigonometric Argument Composite: For $f(x) = \sin(3x)$:

  • $f'(x) = \cos(3x) \cdot \frac{d}{dx}(3x) = 3\cos(3x)$.

  • High-Order Polynomial Composite: For $g(x) = (x^2 + 1)^{100}$:

  • $g'(x) = 100(x^2 + 1)^{99} \cdot \frac{d}{dx}(x^2 + 1) = 100(x^2 + 1)^{99}(2x) = 200x(x^2 + 1)^{99}$.

  • Three-Layer Composite with Exponential and Secant: For $y = \sec(e^{5x})$:

  • Layer 1 (outer): $\sec(u)$, derivative is $\sec(u)\tan(u)$.

  • Layer 2 (middle): $u = e^v$, derivative is $e^v$.

  • Layer 3 (inner): $v = 5x$, derivative is $5$.

  • Step-by-step differentiation:       $\frac{dy}{dx} = \sec(e^{5x})\tan(e^{5x}) \cdot \frac{d}{dx}\left(e^{5x}\right)$       $\frac{dy}{dx} = \sec(e^{5x})\tan(e^{5x}) \cdot e^{5x} \cdot \frac{d}{dx}(5x) = 5 e^{5x} \sec(e^{5x})\tan(e^{5x})$

  • Nested Radicals with Tangent: For $g(u) = \sqrt{u + \sqrt{u + \tan(u)}} = \left[u + (u + \tan(u))^{1/2}\right]^{1/2}$.

  • First outer differentiation:       $g'(u) = \frac{1}{2}\left[u + \sqrt{u + \tan(u)}\right]^{-1/2} \cdot \frac{d}{du}\left(u + [u + \tan(u)]^{1/2}\right)$

  • Inner expression differentiation:       $\frac{d}{du}\left(u + [u + \tan(u)]^{1/2}\right) = 1 + \frac{1}{2}[u + \tan(u)]^{-1/2} \cdot \frac{d}{du}(u + \tan(u))$

  • Innermost differentiation: $\frac{d}{du}(u + \tan(u)) = 1 + \sec^2(u)$.

  • Fully assembled derivative:       $g'(u) = \frac{1}{2}\left[u + \sqrt{u + \tan(u)}\right]^{-1/2} \cdot \left[1 + \frac{1}{2}[u + \tan(u)]^{-1/2}\left(1 + \sec^2(u)\right)\right]$

  • Power of Tangent with Rational Exponential Argument: For $s(t) = \tan^2\left(\frac{2^t + 1}{t}\right)$.

  • Outer derivative: $2\tan\left(\frac{2^t + 1}{t}\right)$.

  • Middle derivative: $\sec^2\left(\frac{2^t + 1}{t}\right)$.

  • Inner derivative via Quotient Rule on $\frac{2^t + 1}{t}$:       $\frac{d}{dt}\left(\frac{2^t + 1}{t}\right) = \frac{t \cdot \frac{d}{dt}(2^t + 1) - (2^t + 1) \cdot \frac{d}{dt}(t)}{t^2} = \frac{t(2^t \ln(2)) - (2^t + 1)(1)}{t^2}$

  • Combined derivative:       $s'(t) = 2\tan\left(\frac{2^t + 1}{t}\right) \sec^2\left(\frac{2^t + 1}{t}\right) \left[\frac{t 2^t \ln(2) - 2^t - 1}{t^2}\right]$

• Graphical and Tabular Evaluation of Composite Derivatives

  • Piecewise Linear Graphical Example: Let $h(x) = f(x^3 - 1)$, where $f$ is defined by the following graph:     

    

  • Differentiating $h(x)$ using the Chain Rule:       $h'(x) = f'(x^3 - 1) \cdot \frac{d}{dx}(x^3 - 1) = f'(x^3 - 1) \cdot 3x^2$

  • Evaluating at $x = 1$:       $h'(1) = f'(1^3 - 1) \cdot 3(1)^2 = f'(0) \cdot 3$

  • Determining $f'(0)$: The value $f'(0)$ represents the slope of the tangent line to $f(x)$ at $x = 0$. From the graph, $f(x)$ is a straight line segment on the interval $[-2, 2]$ connecting $(-2, -1)$, $(0, 0)$, and $(2, 1)$. The slope is:       $m = \frac{1 - 0}{2 - 0} = \frac{1}{2}$

  • Final calculation:       $h'(1) = \frac{1}{2} \cdot 3 = \frac{3}{2}$

  • Tabular Values Example: Let $f$ and $g$ be differentiable functions with values provided in the reference table:     

    

  • Values from the table:

    • At $x = 0$: $f(0) = 1$, $f'(0) = -4$, $g(0) = 2$, $g'(0) = -6$, $g''(0) = 0$.

    • At $x = 1$: $f(1) = 1$, $f'(1) = 4$, $g(1) = -3$, $g'(1) = -3$, $g''(1) = 6$.

    • At $x = 2$: $f(2) = 9$, $f'(2) = 12$, $g(2) = -2$, $g'(2) = 6$, $g''(2) = 12$.

  • Problem: Let $K(x) = g(f(x) + e^{x-1})$. Find $K'(1)$.

  • Differentiating $K(x)$ via Chain Rule:       $K'(x) = g'(f(x) + e^{x-1}) \cdot \frac{d}{dx}\left(f(x) + e^{x-1}\right) = g'(f(x) + e^{x-1}) \cdot \left[f'(x) + e^{x-1}\right]$

  • Evaluating at $x = 1$:       $K'(1) = g'(f(1) + e^{1-1}) \cdot \left[f'(1) + e^{1-1}\right] = g'(f(1) + e^0) \cdot \left[f'(1) + e^0\right]$       $K'(1) = g'(1 + 1) \cdot [4 + 1] = g'(2) \cdot 5 = 6 \cdot 5 = 30$

  • Multi-Function Graphical Analysis:     

    

  • Evaluating derivatives at $x = 1$:

    • For $\frac{d}{dx}\left[\frac{x + g(x)}{f(2x + 1)}\right]$: Apply the Quotient Rule in conjunction with the Chain Rule on the denominator:         $\frac{[1 + g'(x)] f(2x + 1) - [x + g(x)] \cdot f'(2x + 1) \cdot 2}{[f(2x + 1)]^2}$         Evaluate at $x = 1$ using function values and slopes from the graph at $x = 1$ and $2x+1 = 3$.

    • For $\frac{d}{dx}\left[g(f(3x + 1)) + 4^x - g^2(x)\right]$: Apply the Chain Rule across all terms:         $g'(f(3x + 1)) \cdot f'(3x + 1) \cdot 3 + 4^x \ln(4) - 2g(x)g'(x)$         Evaluate at $x = 1$ using coordinates and tangent slopes from the graph.

• Recognizing Derivative Limits via the Chain Rule

  • Problem: Evaluate the limit:     $\lim_{x \to 0} \frac{\sin((3 + x)^2) - \sin(9)}{x}$

  • Limit Definition Formulation: Recall the limit definition of the derivative at a point $a$:     $f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

  • Structural Identification: Replacing the variable name $x$ with $h$:     $\lim_{h \to 0} \frac{\sin((3 + h)^2) - \sin(3^2)}{h}$     This matches the definition of $f'(a)$ with function $f(t) = \sin(t^2)$ evaluated at $a = 3$.

  • Evaluation via Differentiation: Differentiating $f(t) = \sin(t^2)$ using the Chain Rule:     $f'(t) = \cos(t^2) \cdot \frac{d}{dt}(t^2) = 2t \cos(t^2)$

  • Final Value:     $f'(3) = 2(3) \cos(3^2) = 6\cos(9)$     Therefore, $\lim_{x \to 0} \frac{\sin((3 + x)^2) - \sin(9)}{x} = 6\cos(9)$.

Implicit Differentiation

• Implicit vs. Explicit Functions

  • Explicit Functions: A function is explicit when the dependent variable $y$ is isolated directly on one side in terms of the independent variable $x$:     $y = f(x)$

  • Examples: $y = \sqrt{x^3 + 1}$ and $y = x\sin(x)$.

  • Implicit Functions: An equation defines a function implicitly when the relationship between $x$ and $y$ is expressed as a combined relation where $y$ is not explicitly solved:

  • Examples: $x^2 + y^2 = 25$ and $x^3 + y^3 = 6xy$.

  • Multi-Branch Explicit Representation: Solving an implicit equation explicitly may require multiple equations to satisfy the vertical line test.

  • Consider the unit circle $x^2 + y^2 = 1$:       $y^2 = 1 - x^2 \implies y = \pm\sqrt{1 - x^2}$

  • The complete relation is not a single function because it fails the vertical line test:       

    

  • It must be decomposed into two distinct explicit functions:       $\begin{cases} +\sqrt{1 - x^2} & \text{when } y \ge 0 \\ -\sqrt{1 - x^2} & \text{when } y < 0 \end{cases}$       

    

          

    

  • Drawback of Explicit Conversion: Solving for $y$ explicitly often produces cumbersome square roots requiring separate differentiations, which becomes intractable or impossible for higher-degree polynomials (such as $x^3 + y^3 = 6xy$).

• The Implicit Differential Operator Technique

  • Expanding the Differential Operator: The differential operator $\frac{d}{dx}$ can be viewed as multiplying by one via the chain rule differential ratio:     $\frac{d}{dx} \approx \frac{d}{dy} \cdot \frac{dy}{dx}$

  • Fundamental Rule for Differentiating Expressions Containing $y$: When differentiating an expression involving $y$ with respect to $x$, differentiate with respect to $y$ first, then multiply by $\frac{dy}{dx}$ (or $y'$):     $\frac{d}{dx}[g(y)] = \frac{d}{dy}[g(y)] \cdot \frac{dy}{dx} = g'(y) y'$

  • Basic Derivative Demonstrations:

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

  • $\frac{d}{dx}(y^3) = \frac{d}{dy}(y^3) \frac{dy}{dx} = 3y^2 \frac{dy}{dx} = 3y^2 y'$

  • $\frac{d}{dx}(\cos(y)) = \frac{d}{dy}(\cos(y)) \frac{dy}{dx} = -\sin(y) \frac{dy}{dx} = -\sin(y) y'$

  • $\frac{d}{dx}(y^{2/3}) = \frac{d}{dy}(y^{2/3}) \frac{dy}{dx} = \frac{2}{3}y^{-1/3} \frac{dy}{dx} = \frac{2}{3}y^{-1/3} y'$

• Algorithmic Step-by-Step Procedure

  • Step 1: Apply the differentiation operator $\frac{d}{dx}$ to both sides of the equation.

  • Step 2: Differentiate each term, applying the Product, Quotient, and Chain Rules as necessary, appending $y'$ whenever differentiating a term involving $y$.

  • Step 3: Collect all terms containing $y'$ on one side of the equation and move all terms without $y'$ to the opposite side.

  • Step 4: Factor out $y'$ as a common factor.

  • Step 5: Divide by the remaining coefficient factor to solve explicitly for $y'$.

• Comprehensive Implicit Differentiation Examples

  • Example 1: The Circle $x^2 + y^2 = 25$     $\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(25)$     $\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = 0$     $2x + 2y y' = 0$     $2y y' = -2x$     $y' = -\frac{2x}{2y} = -\frac{x}{y}$

  • Example 2: Folium of Descartes $x^3 + y^3 = 6xy$     $\frac{d}{dx}(x^3 + y^3) = \frac{d}{dx}(6xy)$     $3x^2 + 3y^2 y' = 6\left(\frac{d}{dx}(x) \cdot y + x \cdot \frac{d}{dx}(y)\right)$     $3x^2 + 3y^2 y' = 6(1 \cdot y + x y') = 6y + 6x y'$     $3y^2 y' - 6x y' = 6y - 3x^2$     $y'(3y^2 - 6x) = 6y - 3x^2$     $y' = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{3(2y - x^2)}{3(y^2 - 2x)} = \frac{2y - x^2}{y^2 - 2x}$

  • Example 3: Mixed Trigonometric and Product Relation $\sin(x + y) = y^2 \cos(x)$     $\frac{d}{dx}[\sin(x + y)] = \frac{d}{dx}[y^2 \cos(x)]$     $\cos(x + y) \cdot (1 + y') = \left(\frac{d}{dx}(y^2)\right)\cos(x) + y^2\left(\frac{d}{dx}(\cos(x))\right)$     $\cos(x + y) + y' \cos(x + y) = 2y y' \cos(x) - y^2 \sin(x)$     $\cos(x + y) + y^2 \sin(x) = 2y y' \cos(x) - y' \cos(x + y)$     $\cos(x + y) + y^2 \sin(x) = y'[2y \cos(x) - \cos(x + y)]$     $y' = \frac{\cos(x + y) + y^2 \sin(x)}{2y \cos(x) - \cos(x + y)}$

  • Example 4: Second Derivative of $x^4 + y^4 = 16$

  • Find $y'$ first:       $\frac{d}{dx}(x^4 + y^4) = \frac{d}{dx}(16)$       $4x^3 + 4y^3 y' = 0 \implies 4y^3 y' = -4x^3 \implies y' = -\frac{x^3}{y^3}$

  • Differentiate $y'$ with respect to $x$ using the Quotient Rule:       $y'' = \frac{d}{dx}\left(-\frac{x^3}{y^3}\right) = -\frac{\left(\frac{d}{dx}(x^3)\right)y^3 - x^3\left(\frac{d}{dx}(y^3)\right)}{(y^3)^2}$       $y'' = -\frac{3x^2 y^3 - x^3 (3y^2 y')}{y^6} = -\frac{3x^2 y^3 - 3x^3 y^2 y'}{y^6}$

  • Substitute $y' = -\frac{x^3}{y^3}$ into the second derivative expression:       $y'' = -\frac{3x^2 y^3 - 3x^3 y^2 \left(-\frac{x^3}{y^3}\right)}{y^6} = -\frac{3x^2 y^3 + \frac{3x^6 y^2}{y^3}}{y^6} = -\frac{3x^2 y^3 + \frac{3x^6}{y}}{y^6}$       $y'' = -\frac{3x^2 y^4 + 3x^6}{y^7} = -\frac{3x^2(y^4 + x^4)}{y^7}$

  • Simplify using the original constraint $x^4 + y^4 = 16$:       $y'' = -\frac{3x^2(16)}{y^7} = -\frac{48x^2}{y^7}$

  • Example 5: Second-Degree Polynomial Curve $x^2 - 4xy + y^2 = 4$     $\frac{d}{dx}(x^2 - 4xy + y^2) = \frac{d}{dx}(4)$     $2x - 4(1 \cdot y + x y') + 2y y' = 0$     $2x - 4y - 4x y' + 2y y' = 0$     $-4x y' + 2y y' = 4y - 2x$     $y'(-4x + 2y) = 4y - 2x$     $y' = \frac{4y - 2x}{-4x + 2y} = \frac{2(2y - x)}{2(-2x + y)} = \frac{2y - x}{y - 2x}$

  • Example 6: Quartic Implicit Curve $x^4 + x^2 y^2 + y^3 = 5$     $\frac{d}{dx}(x^4 + x^2 y^2 + y^3) = \frac{d}{dx}(5)$     $4x^3 + \left(2x y^2 + x^2 (2y y')\right) + 3y^2 y' = 0$     $4x^3 + 2x y^2 + 2x^2 y y' + 3y^2 y' = 0$     $y'(2x^2 y + 3y^2) = -4x^3 - 2x y^2$     $y' = \frac{-4x^3 - 2x y^2}{2x^2 y + 3y^2}$

  • Example 7: Linear and Trigonometric Mix $\sin(x) + \cos(y) = 2x - 3y$     $\frac{d}{dx}(\sin(x) + \cos(y)) = \frac{d}{dx}(2x - 3y)$     $\cos(x) - \sin(y)y' = 2 - 3y'$     $3y' - \sin(y)y' = 2 - \cos(x)$     $y'(3 - \sin(y)) = 2 - \cos(x)$     $y' = \frac{2 - \cos(x)}{3 - \sin(y)} = \frac{\cos(x) - 2}{\sin(y) - 3}$

  • Example 8: Coupled Exponential Terms $2x e^y + y e^x = 3$     $\frac{d}{dx}(2x e^y + y e^x) = \frac{d}{dx}(3)$     $\left(2 e^y + 2x e^y y'\right) + \left(y' e^x + y e^x\right) = 0$     $2x e^y y' + y' e^x = -2e^y - y e^x$     $y'(2x e^y + e^x) = -(2e^y + y e^x)$     $y' = \frac{-(2e^y + y e^x)}{2x e^y + e^x} = -\frac{2e^y + y e^x}{2x e^y + e^x}$

Introduction to Related Rates and Problem-Solving Strategy

• Core Conceptual Foundation

  • Definition: In a related rates problem, physical or geometric quantities that change over time $t$ are connected through an governing equation. The instantaneous rate of change of one or more quantities is provided, and the objective is to determine the unknown instantaneous rate of change of another quantity at a specific instant.

  • Mathematical Representation: All rates of change are modeled as derivatives with respect to time $t$ ($\frac{d}{dt}$), treating every time-dependent variable as an implicit function of time: $x = x(t)$, $y = y(t)$, $r = r(t)$, $V = V(t)$, $A = A(t)$.

• The Systematic 7-Step Strategy

  • Step 1: Draw and label a diagram. Identify the geometric setup, assign variables to all dynamic quantities that vary with time, and assign constants to dimensions that remain strictly fixed.

  • Step 2: Identify known rates and known values. Express all given quantitative information as numerical values or time derivatives (e.g., $\frac{dx}{dt} = 500\,\text{mph}$).

  • Step 3: Identify the unknown rate. Explicitly state the derivative sought (e.g., $\frac{dz}{dt} = ?$) and specify the precise instant of evaluation (e.g., when $z = 2\,\text{miles}$).

  • Step 4: Write an equation relating the variables. Connect the parameters using geometric, trigonometric, or physical formulas (e.g., Pythagorean theorem, volume, area, similar triangles). Ensure the relation contains only relevant variables.

  • Step 5: Differentiate implicitly with respect to $t$. Apply $\frac{d}{dt}$ across the equation, using the Chain Rule to attach time derivative factors (such as $\frac{dx}{dt}$, $\frac{dy}{dt}$) to each variable.

  • Step 6: Substitute known values. Substitute the instantaneous measurements and rates into the differentiated equation. Critical rule: Never substitute instantaneous values prior to differentiating, as doing so erroneously treats dynamic variables as constants.

  • Step 7: Solve and include units. Algebraically isolate the desired derivative rate and append proper dimensional units (e.g., $\text{cm}^2/\text{s}$, $\text{ft/s}$, $\text{cm/min}$).

• Fundamental Geometric Rate Applications

  • Expanding Square Problem:

  • Geometry: Area of a square of side length $s$ is $A = s^2$.

  • Given: Side length expands at $\frac{ds}{dt} = 6\,\text{cm/s}$.

  • Instant: Find $\frac{dA}{dt}$ when area $A = 16\,\text{cm}^2$.

  • Intermediate Value: $s = \sqrt{16} = 4\,\text{cm}$.

  • Implicit Differentiation with respect to $t$:       $\frac{dA}{dt} = 2s \frac{ds}{dt}$

  • Substitution and Solution:       $\frac{dA}{dt} = 2(4\,\text{cm})(6\,\text{cm/s}) = 48\,\text{cm}^2/\text{s}$

  • Inflating Spherical Balloon Problem:

  • Geometry: Volume of a sphere of radius $r$ is $V = \frac{4}{3}\pi r^3$.

  • Given: Air is pumped into the balloon at a constant rate of $\frac{dV}{dt} = 100\,\text{cm}^3/\text{min}$.

  • Instant: Find the rate of increase of the radius, $\frac{dr}{dt}$, when radius $r = 5\,\text{cm}$.

  • Implicit Differentiation with respect to $t$:       $\frac{dV}{dt} = \frac{4}{3}\pi\left(3r^2 \frac{dr}{dt}\right) = 4\pi r^2 \frac{dr}{dt}$

  • Substitution and Solution:       $100 = 4\pi (5)^2 \frac{dr}{dt} = 100\pi \frac{dr}{dt}$       $\frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi}\,\text{cm/min}$

  • Deflating Spherical Balloon Problem:

  • Geometry: Surface area of a sphere is $A = 4\pi r^2$; diameter is $D = 2r$.

  • Given: Surface area decreases at a rate of $\frac{dA}{dt} = -1\,\text{cm}^2/\text{min}$.

  • Instant: Find the rate of decrease of the diameter, $\frac{dD}{dt}$, when diameter $D = 10\,\text{cm}$ ($r = 5\,\text{cm}$).

  • Differentiating Surface Area with respect to $t$:       $\frac{dA}{dt} = 8\pi r \frac{dr}{dt}$

  • Solving for $\frac{dr}{dt}$:       $-1 = 8\pi(5) \frac{dr}{dt} = 40\pi \frac{dr}{dt} \implies \frac{dr}{dt} = -\frac{1}{40\pi}\,\text{cm/min}$

  • Differentiating Diameter relation $D = 2r$:       $\frac{dD}{dt} = 2 \frac{dr}{dt} = 2\left(-\frac{1}{40\pi}\right) = -\frac{1}{20\pi}\,\text{cm/min}$

  • Conclusion: The diameter decreases at a rate of $\frac{1}{20\pi}\,\text{cm/min}$.

  • Expanding Cube Problem:

  • Geometry: Volume of a cube of edge length $x$ is $V = x^3$.

  • Given: Edge length expands at $\frac{dx}{dt} = 4\,\text{cm/s}$.

  • Instant: Find $\frac{dV}{dt}$ when $x = 15\,\text{cm}$.

  • Implicit Differentiation with respect to $t$:       $\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$

  • Substitution and Solution:       $\frac{dV}{dt} = 3(15)^2(4) = 3(225)(4) = 2700\,\text{cm}^3/\text{s}$

  • Abstract Multi-Variable Function Rate Problem:

  • Given Equation: $y = 3x^4 + 6x$.

  • Rates and State: $x = 1$, $\frac{dx}{dt} = -3$. Find $\frac{dy}{dt}$.

  • Differentiating with respect to $t$:       $\frac{dy}{dt} = 12x^3 \frac{dx}{dt} + 6 \frac{dx}{dt}$

  • Substitution:       $\frac{dy}{dt} = 12(1)^3(-3) + 6(-3) = -36 - 18 = -54$