AP Calculus AB - Derivatives, Implicit Differentiation, Linearization, and Chain Rule

Derivatives of Trigonometric Functions

• Basic Differentiation Formulas for Trigonometric Functions:

  • Derivative of Sine:     $\frac{d}{dx}(\sin(x)) = \cos(x)$

  • Derivative of Cosine:     $\frac{d}{dx}(\cos(x)) = -\sin(x)$

  • Derivative of Tangent:     $\frac{d}{dx}(\tan(x)) = \sec^2(x)$

  • Derivative of Secant:     $\frac{d}{dx}(\sec(x)) = \sec(x)\tan(x)$

  • Derivative of Cotangent:     $\frac{d}{dx}(\cot(x)) = -\csc^2(x)$

  • Derivative of Cosecant:     $\frac{d}{dx}(\csc(x)) = -\csc(x)\cot(x)$

• Example 1: Differentiating Sums of Trigonometric Functions

  • Function: $f(x) = \cos(x) + 2\tan(x)$

  • Applying the derivative term-by-term:     $f'(x) = -\sin(x) + 2\sec^2(x)$     

    

• Finding Tangent Line Equations for Trigonometric Functions:   

  

  • Problem 49: Find the equation of the tangent line to $f(x) = \sin(x) + \cos(x)$ at the point $(0, 1)$.

  • Step 1: Differentiate $f(x)$ to find the slope function:       $f'(x) = \cos(x) - \sin(x)$

  • Step 2: Evaluate slope at $x = 0$:       $f'(0) = \cos(0) - \sin(0) = 1 - 0 = 1$

  • Step 3: Write tangent line equation using slope $m = 1$ and $y$-intercept $b = 1$:       $y = 1x + 1$       $y = x + 1$       

    

  • Problem 50: Find the equation of the tangent line to $f(x) = e^x \cos(x)$ at the point $(0, 1)$.

  • Step 1: Apply the Product Rule to differentiate $f(x)$:       $f'(x) = \frac{d}{dx}(e^x)\cos(x) + e^x \frac{d}{dx}(\cos(x))$       $f'(x) = e^x \cos(x) - e^x \sin(x)$

  • Step 2: Evaluate slope at $x = 0$:       $f'(0) = e^0 \cos(0) - e^0 \sin(0) = 1(1) - 1(0) = 1$

  • Step 3: Write equation of tangent line:       $y = 1x + 1$       $y = x + 1$

The Chain Rule

• Chain Rule Principle and Definition:

  • The Chain Rule calculates the derivative of a composite function $f(g(x))$.

  • Formula:     $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$

  • Leibniz Notation:     $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

• Basic and Trigonometric Chain Rule Examples:

  • Example 1: Differentiating Power of Sine     $\frac{d}{dx}\left(\sin^4(x)\right) = 4\sin^3(x) \cdot \cos(x)$

  • Example 2: Differentiating Trigonometric Function with Inner Linear Argument     $\frac{d}{dx}(\tan(3x)) = 3\sec^2(3x)$

  • Example 3: Cosecant with Inner Linear Argument     $\frac{d}{dx}(\csc(10x)) = -10\csc(10x)\cot(10x)$

  • Example 4: Multi-Layered Chain Rule     $\frac{d}{dx}\left(\sin^4(x^3)\right) = 4\sin^3(x^3) \cdot \cos(x^3) \cdot 3x^2 = 12x^2 \sin^3(x^3)\cos(x^3)$

  • Example 5: Polynomial Power     $\frac{d}{dx}\left((6x + 5)^4\right) = 4(6x + 5)^3 \cdot 6 = 24(6x + 5)^3$

  • Example 6: Single Polynomial Term Raised to Large Power

  • Function: $F(x) = (1 + 2x)^{99}$

  • Derivative: $F'(x) = 99(1 + 2x)^{98} \cdot 2 = 198(1 + 2x)^{98}$

  • Example 7: Composite Polynomial Powers

  • Function: $F(x) = (5x^6 + 2x^3)^4$

  • Derivative: $F'(x) = 4(5x^6 + 2x^3)^3 \cdot (30x^5 + 6x^2)$

  • Example 8: Cosine of Squared Angle

  • Function: $f(\theta) = \cos(\theta^2)$

  • Derivative: $f'(\theta) = -\sin(\theta^2) \cdot 2\theta = -2\theta \sin(\theta^2)$

  • Example 9: Squared Cosine Angle

  • Function: $g(\theta) = \cos^2(\theta) = (\cos\theta)^2$

  • Derivative: $g'(\theta) = 2\cos(\theta) \cdot (-\sin\theta) = -2\sin(\theta)\cos(\theta) = -\sin(2\theta)$

  • Example 10: Product Rule combined with Chain Rule

  • Function: $f(t) = t \sin(\pi t)$

  • Derivative:       $f'(t) = 1 \cdot \sin(\pi t) + t \cdot \cos(\pi t) \cdot \pi$       $f'(t) = \sin(\pi t) + \pi t \cos(\pi t)$

Implicit Differentiation

• Implicit Differentiation Technique and Tangent Line Application:   

  

  • When functions are expressed implicitly (where $y$ is not explicitly isolated), both sides of the equation are differentiated with respect to $x$, treating $y$ as an implicit function of $x$ (applying the chain rule to generate term $\frac{dy}{dx}$).

  • Equation of tangent line using point-slope form:     $y - y_1 = \frac{dy}{dx}(x - x_1)$     

    

• Worked Example: Tangent Line to Complex Implicit Curve

  • Problem 57: Find the equation of the tangent line to $y\sin(2x) = x\cos(2y)$ at the point $\left(\frac{\pi}{2}, \frac{\pi}{4}\right)$.     

    

  • Step 1: Differentiate both sides implicitly with respect to $x$ using the Product and Chain Rules:     $\frac{d}{dx}[y\sin(2x)] = \frac{d}{dx}[x\cos(2y)]$     $\frac{dy}{dx}\sin(2x) + y \cdot 2\cos(2x) = 1 \cdot \cos(2y) + x\left(-\sin(2y) \cdot 2\frac{dy}{dx}\right)$     $\frac{dy}{dx}\sin(2x) + 2y\cos(2x) = \cos(2y) - 2x\sin(2y)\frac{dy}{dx}$

  • Step 2: Substitute point values $x = \frac{\pi}{2}$ and $y = \frac{\pi}{4}$ directly into the differentiated equation:

  • Compute arguments:       $2x = 2\left(\frac{\pi}{2}\right) = \pi$       $2y = 2\left(\frac{\pi}{4}\right) = \frac{\pi}{2}$

  • Trigonometric evaluations:       $\sin(\pi) = 0$       $\cos(\pi) = -1$       $\cos\left(\frac{\pi}{2}\right) = 0$       $\sin\left(\frac{\pi}{2}\right) = 1$

  • Direct substitution:       $\frac{dy}{dx}(0) + 2\left(\frac{\pi}{4}\right)(-1) = 0 - 2\left(\frac{\pi}{2}\right)(1)\frac{dy}{dx}$       $-\frac{\pi}{2} = -\pi \frac{dy}{dx}$       $\frac{dy}{dx} = \frac{-\frac{\pi}{2}}{-\pi} = \frac{1}{2}$

  • Step 3: Write point-slope equation of tangent line:     $y - \frac{\pi}{4} = \frac{1}{2}\left(x - \frac{\pi}{2}\right)$

• Standard Implicit Differentiation Example (Circle Relation):

  • Relation: $x^2 + y^2 = 8$

  • Differentiating both sides:     $2x + 2y\frac{dy}{dx} = 0$     $2y\frac{dy}{dx} = -2x$     $\frac{dy}{dx} = -\frac{x}{y}$

Derivatives of Logarithmic and Exponential Functions

• Natural Logarithm Differentiation Rules:

  • Standard Rule:     $\frac{d}{dx}(\ln(x)) = \frac{1}{x}$

  • Chain Rule with Logarithmic Functions:     $\frac{d}{dx}(\ln(u)) = \frac{1}{u} \cdot \frac{du}{dx}$

• Logarithmic Differentiation Examples:

  • Example 1: Function $f(x) = x\ln(x) - x$     $f'(x) = 1 \cdot \ln(x) + x \cdot \frac{1}{x} - 1$     $f'(x) = \ln(x) + 1 - 1 = \ln(x)$

  • Example 2: Quotient with Natural Logarithm

  • Function: $f(x) = \frac{\ln(x)}{x^2}$

  • Derivative via Quotient Rule:       $f'(x) = \frac{\frac{1}{x} \cdot x^2 - \ln(x) \cdot 2x}{(x^2)^2}$       $f'(x) = \frac{x - 2x\ln(x)}{x^4} = \frac{x(1 - 2\ln(x))}{x^4} = \frac{1 - 2\ln(x)}{x^3}$

  • Example 3: Composite Logarithmic Function Power

  • Function: $f(x) = \sin^2(\ln(x)) = [\sin(\ln(x))]^2$

  • Derivative:       $f'(x) = 2\sin(\ln(x)) \cdot \cos(\ln(x)) \cdot \frac{1}{x}$       $f'(x) = \frac{2\sin(\ln(x))\cos(\ln(x))}{x}$       

    

• Exponential Functions with General Base ($a^x$) and Logarithmic Differentiation:

  • Derivation of $\frac{d}{dx}(2^x)$ via Logarithmic Differentiation:     

    

  • Step 1: Let $y = 2^x$

  • Step 2: Take natural logarithm of both sides:       $\ln(y) = \ln(2^x)$       $\ln(y) = x\ln(2)$

  • Step 3: Differentiate implicitly with respect to $x$:       $\frac{d}{dx}[\ln(y)] = \frac{d}{dx}[x\ln(2)]$       $\frac{1}{y}\frac{dy}{dx} = \ln(2)$

  • Step 4: Solve for $\frac{dy}{dx}$:       $\frac{dy}{dx} = y\ln(2) = 2^x \ln(2)$

  • General Exponential Rule for Base $a$ ($a > 0, a \neq 1$):     $\frac{d}{dx}(a^x) = a^x \ln(a)$     $\frac{d}{dx}(a^{u}) = a^{u}\ln(a) \cdot \frac{du}{dx}$

  • Applied Chain Rule with Base Base $3$ Exponential Function:     

    

  • Function: $h(x) = f(g(x))$ where $f(x) = 3^x$ and $g(x) = x^2 + 4x$

  • Explicit Function: $h(x) = 3^{x^2 + 4x}$

  • Applying Exponential Chain Rule:       $h'(x) = 3^{x^2 + 4x} \cdot (\ln 3) \cdot \frac{d}{dx}(x^2 + 4x)$       $h'(x) = 3^{x^2 + 4x}(\ln 3)(2x + 4)$

Local Linearization and Tangent Line Approximations

• Tangent Line Approximation Concept:

  • The linearization $L(x)$ of a differentiable function $f(x)$ at point $x = a$ provides a linear approximation of the function for values of $x$ near $a$:     $L(x) = f(a) + f'(a)(x - a)$

• Concavity and Over/Underestimation Criteria:   

  

  • Concave Up Criteria ($f''(x) > 0$):

  • The tangent line lies below the graph of $f(x)$.

  • Therefore, the linearization $L(x)$ produces an underestimate ($L(x) < f(x)$).

  • Concave Down Criteria ($f''(x) < 0$):

  • The tangent line lies above the graph of $f(x)$.

  • Therefore, the linearization $L(x)$ produces an overestimate ($L(x) > f(x)$).

• Worked Problem 1: Cubic Function Linearization

  • Find linearization $L(x)$ of $f(x) = x^3 - 2x^2 + 3$ at $a = 2$ and approximate $f(1.9)$.

  • Step 1: Calculate $f(2)$:     $f(2) = (2)^3 - 2(2)^2 + 3 = 8 - 8 + 3 = 3$

  • Step 2: Calculate derivative $f'(x)$ and evaluate $f'(2)$:     $f'(x) = 3x^2 - 4x$     $f'(2) = 3(2)^2 - 4(2) = 12 - 8 = 4$

  • Step 3: Construct linearization equation $L(x)$:     $L(x) = 3 + 4(x - 2)$

  • Step 4: Estimate $f(1.9)$ using $L(1.9)$:     $L(1.9) = 3 + 4(1.9 - 2) = 3 + 4(-0.1) = 3 - 0.4 = 2.6$

  • Step 5: Evaluate concavity to determine error type:     $f''(x) = 6x - 4$     $f''(2) = 6(2) - 4 = 8 > 0$     Since $f''(2) > 0$, $f$ is concave up at $x = 2$. Thus, $L(1.9) = 2.6$ is an underestimate.

• Worked Problem 2: Radical Function Linearization

  • Find linearization $L(x)$ of $f(x) = \sqrt{x}$ centered at $a = 4$ and approximate $f(3.9)$.

  • Step 1: Calculate $f(4)$:     $f(4) = \sqrt{4} = 2$

  • Step 2: Calculate $f'(x)$ and evaluate $f'(4)$:     $f'(x) = \frac{1}{2\sqrt{x}}$     $f'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{4} = 0.25$

  • Step 3: Construct linearization equation $L(x)$:     $L(x) = 2 + 0.25(x - 4)$

  • Step 4: Estimate $f(3.9)$ using $L(3.9)$:     $L(3.9) = 2 + 0.25(3.9 - 4) = 2 + 0.25(-0.1) = 2 - 0.025 = 1.975$

  • Step 5: Evaluate concavity:     $f''(x) = -\frac{1}{4x^{3/2}}$     $f''(4) = -\frac{1}{4(4)^{3/2}} = -\frac{1}{32} < 0$     Since $f''(4) < 0$, $f$ is concave down at $x = 4$. Thus, $L(3.9) = 1.975$ is an overestimate.

• Worked Problem 3: Numerical Estimation of $(1.999)^4$

  • Let $f(x) = x^4$ centered at base point $a = 2$

  • Calculate point values:     $f(2) = 2^4 = 16$     $f'(x) = 4x^3 \implies f'(2) = 4(2)^3 = 32$

  • Linearization equation:     $L(x) = 16 + 32(x - 2)$

  • Evaluate at $x = 1.999$     $L(1.999) = 16 + 32(1.999 - 2) = 16 + 32(-0.001) = 16 - 0.032 = 15.968$

Tabular Differentiation (AP Calculus AB Applications)

• Reference Table of Values for Differentiable Functions $f$ and $g$:

  | $x$ | $f(x)$ | $f'(x)$ | $g(x)$ | $g'(x)$ |   |---|---|---|---|---|   | $1$ | $3$ | $4$ | $2$ | $6$ |   | $2$ | $1$ | $5$ | $8$ | $7$ |   | $3$ | $7$ | $7$ | $2$ | $9$ |

• Composite Function Chain Rule Problems:   

  

  • Problem 78(a): If $h(x) = f(g(x))$, evaluate $h'(1)$.

  • Formula: $h'(x) = f'(g(x)) \cdot g'(x)$

  • Evaluation at $x = 1$:       $h'(1) = f'(g(1)) \cdot g'(1)$       From table: $g(1) = 2$ and $g'(1) = 6$       $h'(1) = f'(2) \cdot 6$       From table: $f'(2) = 5$       $h'(1) = 5 \cdot 6 = 30$

  • Problem 78(b): If H(x) = g(f(x))$, evaluate H'(1).\n - Formula: H'(x) = g'(f(x)) \cdot f'(x)\n - Evaluation at x = 1:\n      H'(1) = g'(f(1)) \cdot f'(1)\n      From table: f(1) = 3$and$f'(1) = 4\n      H'(1) = g'(3) \cdot 4\n      From table: g'(3) = 9\n      H'(1) = 9 \cdot 4 = 36\n\n- Tabular Product Rule Problem:\n  \n - Find \frac{d}{dx}[f(x) \cdot g(x)]\Big|{x=1}\n - Formula:\n    \frac{d}{dx}[f(x)g(x)]\Big|{x=1} = f'(1)g(1) + g'(1)f(1)\n - Substituting values from table:\n    f'(1) = 4, \quad g(1) = 2, \quad g'(1) = 6, \quad f(1) = 3\n    = (4)(2) + (6)(3) = 8 + 18 = 26\n\n- Tabular Quotient Rule Problem:\n  \n - Find \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]\Big|{x=1}\n - Formula:\n    \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]\Big|{x=1} = \frac{f'(1)g(1) - g'(1)f(1)}{(g(1))^2}\n - Substituting values from table:\n    = \frac{(4)(2) - (6)(3)}{(2)^2} = \frac{8 - 18}{4} = \frac{-10}{4} = -\frac{5}{2}$$