19.2 Derivative Definition, Power, Product, Quotient & Transcendental Rules

19.2 Derivative Definition, Power, Product, Quotient & Transcendental Rules

The Derivative as Instantaneous Rate of Change & Tangent Slope

The derivative represents the crowning conceptual achievement of differential calculus, translating average rates of change across finite intervals into instantaneous rates of change at isolated points. Geometrically, consider the secant line passing through two distinct points $(x, f(x))$ and $(x + h, f(x + h))$ on the graph of $y = f(x)$. The slope of this secant line is governed by the difference quotient: $m_{\text{sec}} = \frac{\Delta y}{\Delta x} = \frac{f(x + h) - f(x)}{h}$ As the horizontal increment $h$ approaches zero ($h \to 0$), the secant line pivots around $(x, f(x))$ and converges toward the tangent line. The derivative of $f$ at $x$, denoted $f'(x)$ or $\frac{df}{dx}$, is defined formally as the limit of this difference quotient: $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ provided this limit exists. Alternatively, evaluating the derivative at a specific coordinate $x = a$ can be expressed in point-form: $f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ When this limit converges to a finite real number $L$, $f$ is said to be differentiable at $a$, and $f'(a)$ represents the exact geometric slope $m_{\text{tan}}$ of the line tangent to $y = f(x)$ at $(a, f(a))$.

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Differentiability Implies Continuity & Geometric Failure Modes

A central theorem of calculus establishes an asymmetric relationship between differentiability and continuity:

Theorem: If a function $f$ is differentiable at $x = c$, then $f$ is continuous at $x = c$.

Analytical Verification: To establish continuity, we must demonstrate that $\lim_{x \to c} [f(x) - f(c)] = 0$. Using algebraic manipulation: $\lim_{x \to c} [f(x) - f(c)] = \lim_{x \to c} \left[ \frac{f(x) - f(c)}{x - c} \cdot (x - c) \right] = \left[ \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \right] \cdot \left[ \lim_{x \to c} (x - c) \right] = f'(c) \cdot 0 = 0$ Because $\lim_{x \to c} [f(x) - f(c)] = 0$, it follows that $\lim_{x \to c} f(x) = f(c)$, proving continuity.

The Converse Is Strictly False

Continuity is a necessary condition for differentiability, but it is not sufficient. A function may be completely continuous across an interval yet fail to be differentiable at specific points due to four distinct geometric phenomena:

1. Sharp Corners: A corner occurs where the left-hand and right-hand limits of the difference quotient exist but yield unequal values. The canonical example is $f(x) = |x|$ at $x = 0$: $\lim_{h \to 0^-} \frac{|0 + h| - 0}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1, \quad \lim_{h \to 0^+} \frac{|0 + h| - 0}{h} = \lim_{h \to 0^+} \frac{h}{h} = +1$ Because the one-sided derivatives disagree ($-1 \neq +1$), $f'(0)$ does not exist despite $f$ being continuous everywhere.
2. Cusps: A cusp occurs where the one-sided slopes approach infinite values of opposite signs. For $f(x) = x^{2/3}$ at $x = 0$, $f'(x) = \frac{2}{3x^{1/3}}$; as $x \to 0^-$, the slope approaches $-\infty$, while as $x \to 0^+$, the slope approaches $+\infty$.
3. Vertical Tangents: A vertical tangent occurs where the difference quotient approaches infinity from both sides. For $f(x) = x^{1/3}$ at $x = 0$, $f'(x) = \frac{1}{3x^{2/3}} \to +\infty$ as $x \to 0$. Although continuous and smooth, the slope of the tangent line is undefined ($m = \infty$).
4. Points of Discontinuity: If $f$ is discontinuous at $x = c$, it is automatically non-differentiable at $c$ by contraposition.

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Fundamental Algebraic Differentiation Rules

Rather than evaluating limits of difference quotients repeatedly, calculus establishes algebraic transformation rules:

• Constant Rule: The derivative of any constant $c \in \mathbb{R}$ is zero: $\frac{d}{dx}[c] = 0$
• General Power Rule: For any real constant exponent $n \in \mathbb{R}$: $\frac{d}{dx}[x^n] = n x^{n-1}$ This applies equally to negative powers ($\frac{d}{dx}[x^{-3}] = -3x^{-4}$) and rational exponents ($\frac{d}{dx}[\sqrt[4]{x^3}] = \frac{d}{dx}[x^{3/4}] = \frac{3}{4}x^{-1/4} = \frac{3}{4\sqrt[4]{x}}$).
• Constant Multiple Rule: $\frac{d}{dx}[c f(x)] = c f'(x)$
• Sum and Difference Rules: $\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)$

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The Product and Quotient Rules

When functions combine through multiplication or division, their derivatives cannot be evaluated by differentiating individual factors independently:

• The Product Rule: For differentiable functions $u(x)$ and $v(x)$: $\frac{d}{dx}[u(x) v(x)] = u'(x) v(x) + u(x) v'(x)$
• The Quotient Rule: For differentiable functions $u(x)$ and $v(x)$ with $v(x) \neq 0$: $\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{u'(x) v(x) - u(x) v'(x)}{[v(x)]^2}$ Candidates should guard against the prevalent error of writing $u(x) v'(x) - u'(x) v(x)$ in the numerator; subtraction is non-commutative, so the derivative of the numerator $u'(x)$ must multiply the denominator $v(x)$ first.

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Transcendental Functions: Exponential, Logarithmic & Trigonometric

Transcendental derivatives form a significant component of the FTCE Mathematics 6-12 examination:

Exponential and Logarithmic Rules

• Natural exponential: $\frac{d}{dx}[e^x] = e^x$
• General base exponential ($a > 0, a \neq 1$): $\frac{d}{dx}[a^x] = a^x \ln a$
• Natural logarithm ($x > 0$): $\frac{d}{dx}[\ln x] = \frac{1}{x}$
• General base logarithm ($a > 0, a \neq 1, x > 0$): $\frac{d}{dx}[\log_a x] = \frac{1}{x \ln a}$

The Six Trigonometric Derivatives

The derivatives of the circular functions exhibit an elegant structural symmetry governed by cofunctions: $\frac{d}{dx}[\sin x] = \cos x, \quad \frac{d}{dx}[\cos x] = -\sin x$ $\frac{d}{dx}[\tan x] = \sec^2 x, \quad \frac{d}{dx}[\cot x] = -\csc^2 x$ $\frac{d}{dx}[\sec x] = \sec x \tan x, \quad \frac{d}{dx}[\csc x] = -\csc x \cot x$ Cofunction Negative Sign Mnemonic: The derivative of every trigonometric function beginning with "co-" (cosine, cotangent, cosecant) carries a negative sign, whereas the derivatives of sine, tangent, and secant are positive.

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Higher-Order Derivatives & Kinematics

Differentiating a derivative produces higher-order derivatives:

• Second Derivative: $f''(x) = \frac{d}{dx}[f'(x)] = \frac{d^2y}{dx^2}$, measuring the concavity of $f(x)$ and the rate of change of the slope.
• $n$-th Derivative: $f^{(n)}(x) = \frac{d^n y}{dx^n}$.

In physical kinematics along a straight line, higher-order derivatives model rectilinear motion:

• Position: $s(t)$
• Velocity: $v(t) = s'(t) = \frac{ds}{dt}$
• Acceleration: $a(t) = v'(t) = s''(t) = \frac{d^2s}{dt^2}$
• Jerk: $j(t) = a'(t) = s'''(t) = \frac{d^3s}{dt^3}$

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Comprehensive Differentiation Rules Formula Matrix

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Worked Exemplar: Differentiating a Rational Trigonometric Function

Problem: Let $f(x) = \frac{x^2 \sin x}{e^x}$.

1. Determine the first derivative $f'(x)$ and express it in factored algebraic form.
2. Evaluate $f'(0)$ and $f'(\pi)$.

Step 1: Formulate the Derivative via Product and Quotient Rules. Regard $f(x)$ as a quotient $\frac{u(x)}{v(x)}$ where $u(x) = x^2 \sin x$ and $v(x) = e^x$. First, calculate $u'(x)$ using the Product Rule on the product of $x^2$ and $\sin x$: $u'(x) = \frac{d}{dx}[x^2] \sin x + x^2 \frac{d}{dx}[\sin x] = 2x \sin x + x^2 \cos x$ Next, calculate $v'(x)$: $v'(x) = \frac{d}{dx}[e^x] = e^x$

Step 2: Apply the Quotient Rule. Apply the Quotient Rule formula $\frac{u' v - u v'}{v^2}$: $f'(x) = \frac{(2x \sin x + x^2 \cos x)(e^x) - (x^2 \sin x)(e^x)}{(e^x)^2}$ Factor out the common term $e^x$ from both groupings in the numerator: $f'(x) = \frac{e^x \left[ 2x \sin x + x^2 \cos x - x^2 \sin x \right]}{(e^x)^2}$ Cancel $e^x$ with one factor in the denominator: $f'(x) = \frac{2x \sin x + x^2 \cos x - x^2 \sin x}{e^x}$ Factor common terms in the numerator to produce an optimal factored form: $f'(x) = \frac{x \left[ (2 - x)\sin x + x \cos x \right]}{e^x}$

Step 3: Evaluate at $x = 0$. Substitute $x = 0$ into the simplified derivative: $f'(0) = \frac{0 \cdot \left[ (2 - 0)\sin 0 + 0 \cos 0 \right]}{e^0} = \frac{0 \cdot [2(0) + 0]}{1} = \frac{0}{1} = 0$ This indicates that the curve $y = f(x)$ possesses a horizontal tangent at the origin $(0, 0)$.

Step 4: Evaluate at $x = \pi$. Substitute $x = \pi$ into the simplified derivative: $f'(\pi) = \frac{\pi \left[ (2 - \pi)\sin(\pi) + \pi \cos(\pi) \right]}{e^\pi}$ Because $\sin(\pi) = 0$ and $\cos(\pi) = -1$: $f'(\pi) = \frac{\pi \left[ (2 - \pi)(0) + \pi(-1) \right]}{e^\pi} = \frac{\pi [0 - \pi]}{e^\pi} = \frac{-\pi^2}{e^\pi} = -\pi^2 e^{-\pi}$ Both evaluations are rigorously established.