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

Theoretical Foundations of Differentiation

• The derivative of a function $f(x)$ with respect to $x$ characterizes the instantaneous rate of change of the function and provides the slope of the tangent line to the function's graph at any point $x$.

• The limit definition of the derivative is expressed as:

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

• Differentiation satisfies the property of linearity, which consists of two distinct operational rules:

  • Sum and Difference Rule: The derivative of a sum or difference of two differentiable functions equals the sum or difference of their individual derivatives:

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

• Constant Multiple Rule: The derivative of a constant multiplied by a differentiable function is equal to the constant multiplied by the derivative of the function:

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

• The Power Rule states that for any real constant power $n$:

$\frac{d}{dx}[x^n] = n x^{n - 1}$

• Basic trigonometric differentiation rules establish the derivatives of primary cyclic functions:

$\frac{d}{dx}[\sin(x)] = \cos(x)$

$\frac{d}{dx}[\cos(x)] = -\sin(x)$

$\frac{d}{dx}[\tan(x)] = \sec^2(x)$

Fundamental Derivative Calculations and Algebraic Simplification

• Differentiating linear combinations of power and trigonometric terms:

  • Given the function:

$f(x) = -4x^2 - 4\cos(x)$

• Apply the Constant Multiple Rule and Power Rule to the polynomial term:

$\frac{d}{dx}[-4x^2] = -4(2x) = -8x$

• Apply the Constant Multiple Rule and the trigonometric derivative to the second term:

$\frac{d}{dx}[-4\cos(x)] = -4(-\sin(x)) = 4\sin(x)$

• Combining the differentiated components yields the complete derivative:

$f'(x) = -8x + 4\sin(x)$

• Differentiating rational expressions via monomial division versus the Quotient Rule:

  • Given the rational function:

$f(x) = \frac{x^5 - 9}{x^4}$

• Technique 1: Rewriting using algebraic decomposition prior to differentiation avoids the Quotient Rule and minimizes calculation steps:

$f(x) = \frac{x^5}{x^4} - \frac{9}{x^4} = x - 9x^{-4}$

• Differentiating term-by-term via the Power Rule:

$f'(x) = \frac{d}{dx}[x] - 9 \cdot \frac{d}{dx}[x^{-4}]$

$f'(x) = 1 - 9(-4x^{-5}) = 1 + 36x^{-5}$

• Rewriting with positive exponents or as a single combined rational term:

$f'(x) = 1 + \frac{36}{x^5} = \frac{x^5 + 36}{x^5}$

• Technique 2: Applying the formal Quotient Rule:

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

• Designating $u(x) = x^5 - 9$ and $v(x) = x^4$:

$u'(x) = 5x^4$

$v'(x) = 4x^3$

• Substituting into the formula:

$f'(x) = \frac{(5x^4)(x^4) - (x^5 - 9)(4x^3)}{(x^4)^2}$

$f'(x) = \frac{5x^8 - (4x^8 - 36x^3)}{x^8} = \frac{x^8 + 36x^3}{x^8}$

• Factoring $x^3$ out of the numerator and denominator:

$f'(x) = \frac{x^3(x^5 + 36)}{x^8} = \frac{x^5 + 36}{x^5} = 1 + \frac{36}{x^5}$

Product Rule and Chain Rule Applications

• The Product Rule governs functions defined as products of two distinct factors:

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

• Application of the Product Rule to radical and polynomial factors:

  • Given the function:

$f(u) = \sqrt{u}(5 - u^6)$

• Rewrite the radical in rational exponent notation:

$f(u) = u^{1/2}(5 - u^6)$

• Assign constituent functions:

$g(u) = u^{1/2} \implies g'(u) = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}}$

$h(u) = 5 - u^6 \implies h'(u) = -6u^5$

• Substitute into the Product Rule formula:

$f'(u) = \left(\frac{1}{2}u^{-1/2}\right)(5 - u^6) + u^{1/2}(-6u^5)$

• Multiply and simplify terms:

$f'(u) = \frac{5}{2}u^{-1/2} - \frac{1}{2}u^{11/2} - 6u^{11/2}$

• Combine the like terms with exponent $\frac{11}{2}$:

$-\frac{1}{2}u^{11/2} - \frac{12}{2}u^{11/2} = -\frac{13}{2}u^{11/2}$

$f'(u) = \frac{5}{2\sqrt{u}} - \frac{13}{2}u^{11/2}$

• Alternative single-fraction form:

$f'(u) = \frac{5 - u^6 - 12u^6}{2\sqrt{u}} = \frac{5 - 13u^6}{2\sqrt{u}}$

• Combining Product Rule and Trigonometric Differentiation:

  • Given the multi-term function:

$f(s) = 9s \sin(s) + 5\cos(s)$

• The first term, $9s \sin(s)$, requires the Product Rule where $u(s) = 9s$ and $v(s) = \sin(s)$:

$\frac{d}{ds}[9s \sin(s)] = \frac{d}{ds}[9s] \cdot \sin(s) + 9s \cdot \frac{d}{ds}[\sin(s)] = 9\sin(s) + 9s\cos(s)$

• The second term, $5\cos(s)$, differentiates directly:

$\frac{d}{ds}[5\cos(s)] = -5\sin(s)$

• Summing the differentiated parts:

$f'(s) = 9\sin(s) + 9s\cos(s) - 5\sin(s)$

• Combine the like trigonometric terms ($9\sin(s) - 5\sin(s)$):

$f'(s) = 4\sin(s) + 9s\cos(s)$

• The Chain Rule provides the derivative for composite functions $f(t) = g(h(t))$:

$f'(t) = g'(h(t)) \cdot h'(t)$

• General Power Rule (composite power function):

$\frac{d}{dt}\left[[u(t)]^n\right] = n[u(t)]^{n - 1} \cdot u'(t)$

• Differentiating a high-degree composite polynomial expression:

  • Given the function:

$f(t) = (4 + 7t^5)^8$

• Define the inner function and the outer function:

$u(t) = 4 + 7t^5$

$g(u) = u^8$

• Compute derivatives:

$g'(u) = 8u^7$

$u'(t) = \frac{d}{dt}[4 + 7t^5] = 35t^4$

• Multiply via the Chain Rule:

$f'(t) = 8(4 + 7t^5)^7 \cdot (35t^4)$

• Multiply the constant coefficients ($8 \times 35 = 280$):

$f'(t) = 280t^4 (4 + 7t^5)^7$

Implicit Differentiation

• When an equation defines a relation between $x$ and $y$ implicitly rather than explicitly as $y = f(x)$, implicit differentiation is performed by differentiating each term with respect to $x$, treating $y$ as a differentiable function of $x$, and applying the Chain Rule (multiplying by $\frac{dy}{dx}$ whenever differentiating a function of $y$).

• Derivation of $\frac{dy}{dx}$ for the relation $x \cos(y) = 1$:

  • Differentiate both sides of the equation with respect to $x$:

$\frac{d}{dx}[x \cos(y)] = \frac{d}{dx}[1]$

• Apply the Product Rule on the left-hand side, where the first factor is $x$ and the second factor is $\cos(y)$:

$\frac{d}{dx}[x] \cdot \cos(y) + x \cdot \frac{d}{dx}[\cos(y)] = 0$

• Compute the derivative of $\cos(y)$ with respect to $x$ using the Chain Rule:

$\frac{d}{dx}[\cos(y)] = -\sin(y) \cdot \frac{dy}{dx}$

• Substitute back into the product expression:

$1 \cdot \cos(y) + x\left(-\sin(y) \frac{dy}{dx}\right) = 0$

$\cos(y) - x\sin(y) \frac{dy}{dx} = 0$

• Isolate the term containing $\frac{dy}{dx}$:

$x\sin(y) \frac{dy}{dx} = \cos(y)$

• Solve algebraically for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{\cos(y)}{x\sin(y)} = \frac{\cot(y)}{x}$

• Alternative representations using the original identity $x = \frac{1}{\cos(y)} = \sec(y)$:

$\frac{dy}{dx} = \frac{\cos(y)}{\sec(y)\sin(y)} = \frac{\cos^2(y)}{\sin(y)} = \cot(y)\cos(y)$

Related Rates: Principles and Methodology

• Related rates problems involve finding the rate at which one physical quantity changes by relating it to other quantities whose rates of change are known.

• Systematic Problem-Solving Procedure:

  • Step 1: Identify and assign variables to all known quantities and unknown rates changing over time $t$.

  • Step 2: Establish a geometric or physical relation connecting the variables, independent of time.

  • Step 3: Eliminate unwanted auxiliary variables using geometric constraints, such as similar triangles or substitution, to express the equation in terms of the desired variable.

  • Step 4: Differentiate both sides implicitly with respect to time $t$ using the Chain Rule.

  • Step 5: Substitute specific snapshot values into the differentiated equation and solve for the unknown rate.

Related Rates Problem 1: Inverted Conical Reservoir

• Physical Description and Given Constants:

  • An inverted conical tank has its vertex pointing downward.

  • Total height of the tank: $H = 60\,\text{ft}$.

  • Total base radius at the top rim: $R = 18\,\text{ft}$.

  • Volumetric outflow rate: Water drains out at a constant rate of $9\,\text{ft}^3/\text{min}$, meaning:

$\frac{dV}{dt} = -9\,\text{ft}^3/\text{min}$

• Specified snapshot water depth: $h = 40\,\text{ft}$.

• Target rate: The rate at which the depth of the water is decreasing ($\frac{dh}{dt}$).

  • Geometric Modeling via Similar Triangles:

• The cross-section of the cone creates similar right triangles between the full tank dimensions and the water cone dimensions at any time $t$:

$\frac{r}{h} = \frac{R}{H} = \frac{18}{60} = \frac{3}{10}$

• Express radius $r$ strictly in terms of depth $h$ to eliminate $r$:

$r = \frac{3}{10}h$

• Volume Equation Formulation:

  • The formula for the volume of a circular cone is:

$V = \frac{1}{3}\pi r^2 h$

• Substitute $r = \frac{3}{10}h$ into the volume equation:

$V = \frac{1}{3}\pi \left(\frac{3}{10}h\right)^2 h = \frac{1}{3}\pi \left(\frac{9}{100}h^2\right) h$

$V = \frac{3\pi}{100}h^3$

• Differentiation with Respect to Time:

  • Differentiating both sides with respect to $t$ via the Chain Rule:

$\frac{dV}{dt} = \frac{3\pi}{100} \cdot \left(3h^2 \frac{dh}{dt}\right) = \frac{9\pi}{100}h^2 \frac{dh}{dt}$

• Evaluation at the Snapshot State ($h = 40\,\text{ft}$, $\frac{dV}{dt} = -9\,\text{ft}^3/\text{min}$):

$-9 = \frac{9\pi}{100}(40)^2 \frac{dh}{dt}$

$(40)^2 = 1600$

$-9 = \frac{9\pi}{100}(1600) \frac{dh}{dt}$

$-9 = 144\pi \frac{dh}{dt}$

• Solving for $\frac{dh}{dt}$:

$\frac{dh}{dt} = \frac{-9}{144\pi} = -\frac{1}{16\pi}\,\text{ft/min}$

• Interpretation of the Result:

  • The negative sign indicates that the water level is dropping.

  • The depth of the water is decreasing at a rate of $\frac{1}{16\pi}\,\text{ft/min}$ (approximately $0.01989\,\text{ft/min}$).

Related Rates Problem 2: Sliding Ladder

• Physical Description and Given Constants:

  • A rigid ladder of fixed length $L = 25\,\text{ft}$ leans against a vertical wall, with its foot on horizontal ground.

  • The top of the ladder slips downward along the wall at a speed of $1\,\text{ft/s}$.

  • Letting $y(t)$ denote the vertical height of the ladder's top on the wall:

$\frac{dy}{dt} = -1\,\text{ft/s}$

• Letting $x(t)$ denote the horizontal distance from the base of the wall to the foot of the ladder:

• Specified snapshot position: $x = 7\,\text{ft}$.

• Target rate: The rate at which the bottom of the ladder is moving away from the wall ($\frac{dx}{dt}$).

  • Geometric Modeling via the Pythagorean Theorem:

• The wall, the ground, and the ladder form a right triangle at all times:

$x^2 + y^2 = L^2$

$x^2 + y^2 = 25^2 = 625$

• Determining the Instantaneous Height $y$:

  • When $x = 7\,\text{ft}$:

$7^2 + y^2 = 625$

$49 + y^2 = 625$

$y^2 = 625 - 49 = 576$

$y = \sqrt{576} = 24\,\text{ft}$

• This configuration corresponds to the standard $7-24-25$ integer Pythagorean triple.

  • Implicit Differentiation with Respect to Time:

• Differentiating the Pythagorean relation with respect to $t$:

$\frac{d}{dt}[x^2 + y^2] = \frac{d}{dt}[625]$

$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$

• Dividing the entire equation by 2:

$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$

• Evaluation at the Snapshot State ($x = 7\,\text{ft}$, $y = 24\,\text{ft}$, $\frac{dy}{dt} = -1\,\text{ft/s}$):

$7 \frac{dx}{dt} + 24(-1) = 0$

$7 \frac{dx}{dt} - 24 = 0$

$7 \frac{dx}{dt} = 24$

$\frac{dx}{dt} = \frac{24}{7}\,\text{ft/s}$

• Interpretation of the Result:

  • The positive sign indicates that distance $x$ is increasing over time.

  • The bottom of the ladder is slipping along the ground away from the wall at a speed of $\frac{24}{7}\,\text{ft/s}$ (approximately $3.4286\,\text{ft/s}$).