6.1 Definition of Derivative, Power, Product, Quotient & Chain Rules

Definition of Derivative, Power, Product, Quotient & Chain Rules

Differential calculus is the foundational engine of intermediate mathematics (FSc Part II) and represents one of the heaviest-tested topics on the Pakistan Engineering Colleges Admission Test (ECAT) conducted by UET Lahore. With 30 mathematics questions to answer in approximately 30 minutes, candidates must move beyond lengthy multi-step algebraic expansions and master rapid, systematic differentiation. This section establishes the rigorous definition of the derivative from first principles, examines differentiability versus continuity, derives the core operational rules of calculus, and presents high-speed shortcuts tailored for the ECAT examination.

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1. The Derivative from First Principles (Ab Initio Method)

Analytical Definition

Let $y = f(x)$ be a real-valued function defined in an open interval containing $x$. Let $x$ receive a small increment $\delta x$ (or $\Delta x$), producing a corresponding increment $\delta y = f(x + \delta x) - f(x)$ in the dependent variable $y$.

The average rate of change over the interval $[x, x + \delta x]$ is given by the difference quotient:

$\frac{\delta y}{\delta x} = \frac{f(x + \delta x) - f(x)}{\delta x}$

The derivative of $f$ with respect to $x$, denoted by $f'(x)$ or $\frac{dy}{dx}$, is defined as the limit of this difference quotient as $\delta x$ approaches zero, provided this limit exists:

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

Alternative standard notations introduced by historical mathematicians include:

• Leibniz notation: $\frac{dy}{dx}$ or $\frac{d}{dx}[f(x)]$
• Lagrange notation: $f'(x)$ or $y'$
• Cauchy notation: $D f(x)$ or $D_x y$
• Newton notation: $\dot{y}$ (primarily used in mechanics for time derivatives)

The Four-Step Ab Initio Procedure

In textbook exercises, evaluating derivatives from first principles follows four systematic steps:

1. Increment: Compute $y + \delta y = f(x + \delta x)$.
2. Difference: Subtract $y = f(x)$ to isolate $\delta y = f(x + \delta x) - f(x)$.
3. Quotient: Divide both sides by $\delta x$ to form $\frac{\delta y}{\delta x}$.
4. Limit: Take $\lim_{\delta x \to 0} \frac{\delta y}{\delta x}$ to evaluate $f'(x)$.

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2. Geometric and Physical Interpretations

Geometric Interpretation: Slope of the Tangent Line

Consider the secant line passing through two points $P(x, f(x))$ and $Q(x + \delta x, f(x + \delta x))$ on the graph of $y = f(x)$. The slope of the secant line $PQ$ is:

$m_{\text{sec}} = \frac{f(x + \delta x) - f(x)}{\delta x}$

As $\delta x \to 0$, point $Q$ moves along the curve toward $P$. The secant line $PQ$ rotates into a limiting position, which is the tangent line to the curve at point $P$. Therefore:

$m_{\text{tan}} = \lim_{\delta x \to 0} m_{\text{sec}} = f'(x)$

The derivative $f'(x_0)$ evaluates the exact numerical gradient (slope $m = \tan\theta$) of the tangent line to the curve at $x = x_0$.

Physical Interpretation: Instantaneous Rate of Change

If $s(t)$ represents the displacement of a particle along a straight line at time $t$, the average velocity during the time interval $[t, t + \Delta t]$ is $\frac{\Delta s}{\Delta t}$. The instantaneous velocity at time $t$ is the derivative:

$v(t) = \lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t} = \frac{ds}{dt}$

Similarly, instantaneous acceleration is the derivative of velocity: $a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}$. In general, if quantity $y$ depends on $x$, $\frac{dy}{dx}$ quantifies the instantaneous rate of change of $y$ per unit change in $x$.

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3. Differentiability vs. Continuity

A function $f(x)$ is differentiable at $x = c$ if the bilateral derivative exists at $c$, meaning the left-hand derivative equals the right-hand derivative:

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

Fundamental Theorem

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

Proof: For $x \neq c$, write the algebraic identity: $f(x) - f(c) = \frac{f(x) - f(c)}{x - c} \cdot (x - c)$ Taking the limit as $x \to c$: $\lim_{x \to c} [f(x) - f(c)] = \lim_{x \to c} \left[ \frac{f(x) - f(c)}{x - c} \right] \cdot \lim_{x \to c} (x - c) = f'(c) \cdot 0 = 0$ $\lim_{x \to c} f(x) = f(c)$ Hence, $f$ is continuous at $c$.

The Converse Fails: Continuity Does Not Imply Differentiability

A function can be continuous everywhere while failing to be differentiable at specific points. Non-differentiability occurs under three primary geometric conditions:

1. Sharp Corner / Node: The left-hand and right-hand tangents have different finite slopes. For example, $f(x) = |x|$ at $x = 0$ is continuous, but $f'_-(0) = -1$ while $f'_+(0) = +1$. Because $-1 \neq +1$, $f'(0)$ does not exist.
2. Cusp: Slopes of secant lines approach $+\infty$ on one side and $-\infty$ on the other (e.g., $f(x) = x^{2/3}$ at $x = 0$).
3. Vertical Tangent: The slope approaches $\pm\infty$ continuously (e.g., $f(x) = x^{1/3}$ at $x = 0$, where $f'(x) = \frac{1}{3x^{2/3}} \to \infty$).

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4. Fundamental Algebraic Differentiation Rules

1. The Constant Rule

If $c$ is a constant real number and $f(x) = c$, then: $\frac{d}{dx}(c) = 0$

2. The Power Rule

For any real constant $n \in \mathbb{R}$: $\frac{d}{dx}(x^n) = n x^{n-1}$

Proof for rational exponent $n = \frac{p}{q}$: Let $y = x^{p/q} \implies y^q = x^p$. Differentiating implicitly yields $q y^{q-1} \frac{dy}{dx} = p x^{p-1}$, so $\frac{dy}{dx} = \frac{p}{q} \frac{x^{p-1}}{(x^{p/q})^{q-1}} = \frac{p}{q} x^{(p-1) - p(1 - 1/q)} = \frac{p}{q} x^{p/q - 1} = n x^{n-1}$.

Common instances frequently used on the ECAT:

• $\frac{d}{dx}(x) = 1$
• $\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$
• $\frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2}$
• $\frac{d}{dx}\left(\frac{1}{\sqrt{x}}\right) = -\frac{1}{2x^{3/2}}$

3. The Constant Multiple Rule

If $c$ is a constant and $u(x)$ is differentiable: $\frac{d}{dx}[c \cdot u(x)] = c \cdot \frac{du}{dx}$

4. The Sum and Difference Rule

If $u(x)$ and $v(x)$ are differentiable functions: $\frac{d}{dx}[u(x) \pm v(x)] = \frac{du}{dx} \pm \frac{dv}{dx}$

5. The Leibniz Product Rule

The derivative of the product of two differentiable functions $u(x)$ and $v(x)$ is: $\frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx} = u v' + v u'$

For a product of three differentiable functions $u, v, w$: $\frac{d}{dx}[u v w] = u' v w + u v' w + u v w'$

6. The Quotient Rule

If $u(x)$ and $v(x)$ are differentiable and $v(x) \neq 0$: $\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} = \frac{u' v - u v'}{v^2}$

ECAT Trap Alert: A frequent candidate error is reversing the numerator terms to write $u v' - u' v$. The quotient rule strictly requires the denominator $v$ multiplied by the derivative of the numerator $u'$ first: $\text{Numerator} = (\text{Bottom}) \cdot (\text{Top}') - (\text{Top}) \cdot (\text{Bottom}')$. Reversing this introduces an erroneous minus sign.

7. The Chain Rule for Composite Functions

If $y = f(u)$ is differentiable with respect to $u$, and $u = g(x)$ is differentiable with respect to $x$, the composite function $y = (f \circ g)(x) = f(g(x))$ has derivative: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = f'(g(x)) \cdot g'(x)$

For multi-stage composite functions $y = f(u)$, $u = g(v)$, $v = h(x)$: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$

8. Generalized Power Rule

Applying the chain rule to $[f(x)]^n$ yields: $\frac{d}{dx}\left([f(x)]^n\right) = n [f(x)]^{n-1} \cdot f'(x)$

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5. Derivative Rules Summary Table

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6. High-Speed ECAT Differentiation Shortcuts

Under the 60-second-per-question constraint of the ECAT, applying standard quotient or chain rule expansions directly is often too slow. The following algebraic shortcuts allow instant evaluations.

Shortcut 1: Derivative of Linear Fractional (Homographic) Functions

Consider the linear rational expression: $y = \frac{ax + b}{cx + d}$ Using the standard quotient rule: $\frac{dy}{dx} = \frac{a(cx + d) - (ax + b)c}{(cx + d)^2} = \frac{acx + ad - acx - bc}{(cx + d)^2} = \frac{ad - bc}{(cx + d)^2}$

ECAT Memory Aid: The numerator is simply the $2 \times 2$ determinant of the coefficients: $\frac{d}{dx}\left(\frac{ax + b}{cx + d}\right) = \frac{\begin{vmatrix} a & b \\ c & d \end{vmatrix}}{(cx + d)^2} = \frac{ad - bc}{(cx + d)^2}$

Example: Find $\frac{d}{dx}\left(\frac{4x - 7}{3x + 5}\right)$. $\text{Numerator} = (4)(5) - (-7)(3) = 20 + 21 = 41 \implies \frac{dy}{dx} = \frac{41}{(3x + 5)^2}$ Solved in 4 seconds without writing scratch calculations!

Shortcut 2: Fast Radical Differentiation

For any nested square root $y = \sqrt{f(x)}$: $\frac{dy}{dx} = \frac{f'(x)}{2\sqrt{f(x)}}$ For reciprocal roots $y = \frac{1}{\sqrt{f(x)}} = [f(x)]^{-1/2}$: $\frac{dy}{dx} = -\frac{f'(x)}{2[f(x)]^{3/2}} = -\frac{f'(x)}{2 f(x) \sqrt{f(x)}}$

Shortcut 3: Reciprocal Power Shortcut

For functions of the form $y = \frac{1}{[f(x)]^k}$: $\frac{dy}{dx} = -\frac{k \cdot f'(x)}{[f(x)]^{k+1}}$