Differentiability
Differentiability & Methods of Differentiation
Foundations of Differentiability
Definition and Geometric InterpretationTopic 1
A real-valued function $f(x)$ defined on an open interval $(a, b)$ is said to be differentiable at a point $c \in (a, b)$ if the limit of the difference quotient exists finitely. Mathematically, the derivative $f'(c)$ is defined as: \[ f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} \] Geometrically, the difference quotient $\frac{f(c+h)-f(c)}{h}$ represents the slope of the secant line passing through the points $P(c, f(c))$ and $Q(c+h, f(c+h))$ on the curve. As $h \to 0$, the point $Q$ moves along the curve towards $P$. If the limit exists, the secant line approaches a limiting position, which is defined as the tangent line to the curve at $P$. Therefore, $f'(c)$ represents the exact slope of the tangent line to the curve at $x = c$.
For a function to be differentiable at $x=c$, the direction of approach from either side must yield matching slopes. We define:- Left-Hand Derivative (LHD): $f'(c^-) = \lim_{h \to 0^+} \frac{f(c-h) - f(c)}{-h}$
- Right-Hand Derivative (RHD): $f'(c^+) = \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h}$
Examine the differentiability of the function $f(x) = x|x|$ at $x = 0$.
Show solution
- Step 1: Evaluate Right-Hand Derivative (RHD)
Using $c = 0$, we have $f(0) = 0$: \[ \text{RHD} = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h^2 - 0}{h} = \lim_{h \to 0^+} h = 0 \] - Step 2: Evaluate Left-Hand Derivative (LHD)
Using the definition for the left side: \[ \text{LHD} = \lim_{h \to 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h \to 0^+} \frac{-( -h )^2 - 0}{-h} = \lim_{h \to 0^+} \frac{-h^2}{-h} = \lim_{h \to 0^+} h = 0 \]
Continuity vs. Differentiability and Breakpoint FailuresTopic 2
A fundamental theorem of calculus establishes that differentiability implies continuity. If a function $f(x)$ is differentiable at $x = c$, it is guaranteed to be continuous at $x = c$. \[ \text{Proof: } \lim_{x \to c} (f(x) - f(c)) = \lim_{x \to c} \left[ \frac{f(x) - f(c)}{x - c} \cdot (x - c) \right] = f'(c) \cdot 0 = 0 \implies \lim_{x \to c} f(x) = f(c) \] However, the converse is strictly false. A function can be perfectly continuous at a point but fail to be differentiable. The classic counterexample is $f(x) = |x|$ at $x = 0$. It forms a continuous V-shape path, but $\text{LHD} = -1 \neq 1 = \text{RHD}$.
Points of non-differentiability on a continuous path are structurally classified into:- Corners: Sudden shifts in direction where both LHD and RHL exist but are unequal (e.g., $f(x) = |x|$ at $x=0$).
- Cusps: Sharp points where the one-sided derivatives approach opposite infinities ($\infty$ and $-\infty$), such as $f(x) = x^{2/3}$ at $x = 0$.
- Vertical Tangents: Points where both one-sided derivatives approach the same infinity ($\pm\infty$), making the tangent line perfectly vertical (e.g., $f(x) = x^{1/3}$ at $x = 0$).
- Discontinuities: Any break, hole, or jump automatically destroys differentiability.
Examine the differentiability of $f(x) = \max(x, x^2)$ across the real line and identify points of failure.
Show solution
- At $x = 0$:
$\text{LHD} = \frac{d}{dx}(x) = 1$.
$\text{RHD} = \frac{d}{dx}(x^2)\Big|_{x=0} = 2(0) = 0$.
Since $\text{LHD} \neq \text{RHD}$ ($1 \neq 0$), $x = 0$ is a corner point. - At $x = 1$:
$\text{LHD} = \frac{d}{dx}(x^2)\Big|_{x=1} = 2(1) = 2$.
$\text{RHD} = \frac{d}{dx}(x) = 1$.
Since $\text{LHD} \neq \text{RHD}$ ($2 \neq 1$), $x = 1$ is a corner point.
Analytical Differentiation & Core Rules
Derivatives of Standard Functions and Operational AlgebraTopic 1
- Product Rule (Leibnitz Rule): $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$
- Quotient Rule: $\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$
- Chain Rule: For composite maps $y = f(g(x))$, let $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = f'(g(x)) \cdot g'(x)$. This generalizes across multiple nested layers.
Differentiate $y = e^{\sin(x^2)}$ with respect to $x$.
Show solution
- Layer 1: Exponential function structure $\frac{d}{du}(e^u) = e^u$, where $u = \sin(x^2)$. \[ \frac{dy}{dx} = e^{\sin(x^2)} \cdot \frac{d}{dx}(\sin(x^2)) \]
- Layer 2: Trigonometric wrapper $\frac{d}{dv}(\sin v) = \cos v$, where $v = x^2$. \[ \frac{d}{dx}(\sin(x^2)) = \cos(x^2) \cdot \frac{d}{dx}(x^2) \]
- Layer 3: Algebraic interior power $\frac{d}{dx}(x^2) = 2x$.
Advanced Operational TechniquesTopic 2
- Implicit Differentiation: When equations bind variables implicitly as $f(x, y) = 0$, we differentiate all terms with respect to $x$, treating $y$ as a function of $x$ (applying the chain rule to generate $\frac{dy}{dx}$ terms), and then algebraically isolate $\frac{dy}{dx}$.
- Logarithmic Differentiation: Essential for expressions of the form $y = [f(x)]^{g(x)}$ or massive products. Taking the natural log transforms exponents into products: $\ln y = g(x) \ln[f(x)]$. Differentiating yields $\frac{1}{y}\frac{dy}{dx} = g'(x)\ln[f(x)] + g(x)\frac{f'(x)}{f(x)}$.
- Parametric Differentiation: If $x = f(t)$ and $y = g(t)$, then the derivative is calculated as the ratio of their independent rates of change over the parameter: \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)} \quad (\text{provided } f'(t) \neq 0) \]
Find $\frac{dy}{dx}$ if $x^y = e^{x-y}$.
Show solution
Given variable expressions in both the base and exponent positions, we apply logarithmic transformation directly: \[ \ln(x^y) = \ln(e^{x-y}) \implies y \ln x = x - y \] Isolate $y$ explicitly to streamline calculation: \[ y \ln x + y = x \implies y(\ln x + 1) = x \implies y = \frac{x}{1 + \ln x} \] Now compute the derivative using the quotient rule: \[ \frac{dy}{dx} = \frac{(1)(1 + \ln x) - (x)\left(\frac{1}{x}\right)}{(1 + \ln x)^2} = \frac{1 + \ln x - 1}{(1 + \ln x)^2} = \frac{\ln x}{(1 + \ln x)^2} \] Final Answer: $\frac{dy}{dx} = \frac{\ln x}{(1 + \ln x)^2}$.
Higher-Order Calculus & Special Systems
Inverse Trigonometric Derivatives and Parametric ExtensionsTopic 1
Derivatives of inverse trigonometric mappings require strict adherence to standard domain restrictions: \[ \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}, \quad \frac{d}{dx}(\cos^{-1} x) = -\frac{1}{\sqrt{1-x^2}} \quad (\text{for } |x| < 1) \] \[ \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}, \quad \frac{d}{dx}(\cot^{-1} x) = -\frac{1}{1+x^2} \quad (\text{for } x \in \mathbb{R}) \]
When extending parametric models to the second derivative $\frac{d^2y}{dx^2}$, a common mistake is simply dividing $\frac{d^2y}{dt^2}$ by $\frac{d^2x}{dt^2}$. The correct approach requires applying the chain rule to the first derivative with respect to $x$: \[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} \]
If $x = a(\theta - \sin \theta)$ and $y = a(1 - \cos \theta)$, determine $\frac{d^2y}{dx^2}$ at $\theta = \pi/2$.
Show solution
First, evaluate the individual rates of change over the parameter $\theta$: \[ \frac{dx}{dt} = a(1 - \cos \theta), \quad \frac{dy}{dt} = a \sin \theta \implies \frac{dy}{dx} = \frac{a \sin \theta}{a(1 - \cos \theta)} = \frac{2 \sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2 \sin^2\frac{\theta}{2}} = \cot\frac{\theta}{2} \] Now differentiate $\frac{dy}{dx}$ with respect to $\theta$ and scale by $\frac{dt}{dx}$: \[ \frac{d}{d\theta}\left(\frac{dy}{dx}\right) = \frac{d}{d\theta}\left(\cot\frac{\theta}{2}\right) = -\frac{1}{2}\csc^2\frac{\theta}{2} \] \[ \frac{d^2y}{dx^2} = \frac{-\frac{1}{2}\csc^2\frac{\theta}{2}}{a(1 - \cos \theta)} = \frac{-\frac{1}{2}\csc^2\frac{\theta}{2}}{2a \sin^2\frac{\theta}{2}} = -\frac{1}{4a}\csc^4\frac{\theta}{2} \] Evaluate this expression at the target point $\theta = \frac{\pi}{2} \implies \frac{\theta}{2} = \frac{\pi}{4}$: \[ \frac{d^2y}{dx^2}\Big|_{\theta=\pi/2} = -\frac{1}{4a} ( \sqrt{2} )^4 = -\frac{4}{4a} = -\frac{1}{a} \] Final Answer: $-\frac{1}{a}$.
Matrix Differentiation and Leibnitz Successive ExtensionsTopic 2
- Differentiation of Determinants: To differentiate a determinant, differentiate one row (or column) at a time, keeping the remaining rows (or columns) unaltered, and sum the resulting determinants: \[ \frac{d}{dx} \begin{vmatrix} f_1(x) & g_1(x) \\ f_2(x) & g_2(x) \end{vmatrix} = \begin{vmatrix} f_1'(x) & g_1'(x) \\ f_2(x) & g_2(x) \end{vmatrix} + \begin{vmatrix} f_1(x) & g_1(x) \\ f_2'(x) & g_2'(x) \end{vmatrix} \]
- Leibnitz Theorem for $n$-th Derivatives: Generalizes the product rule to higher-order calculations using binomial coefficients: \[ d^n(uv) = \sum_{k=0}^n \binom{n}{k} u^{(n-k)} v^{(k)} \]
Evaluate $F'(0)$ if $F(x) = \begin{vmatrix} \ln(1+x) & x \\ e^x & x^2 + 1 \end{vmatrix}$.
Show solution
Apply the determinant differentiation rule: \[ F'(x) = \begin{vmatrix} \frac{1}{1+x} & 1 \\ e^x & x^2 + 1 \end{vmatrix} + \begin{vmatrix} \ln(1+x) & x \\ e^x & 2x \end{vmatrix} \] Evaluate each component directly at $x = 0$: \[ F'(0) = \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} + \begin{vmatrix} 0 & 0 \\ 1 & 0 \end{vmatrix} = (1 - 1) + (0 - 0) = 0 \] Final Answer: $0$.
Mean Value Theorems & Series Expansions
Rolle's Theorem and Lagrange's VariationsTopic 1
- Rolle's Theorem: Let $f:[a, b] \to \mathbb{R}$ be a function that satisfies:
- Continuous on the closed interval $[a, b]$
- Differentiable on the open interval $(a, b)$
- $f(a) = f(b)$
- Lagrange's Mean Value Theorem (LMVT): Drops the endpoint equality constraint. If conditions (i) and (ii) hold, there exists at least one $c \in (a, b)$ such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \] Geometrically, the tangent line at $x=c$ is parallel to the secant line connecting the endpoints $(a, f(a))$ and $(b, f(b))$.
Verify Rolle's Theorem for $f(x) = x(x-1)(x-2)$ on $[0, 2]$.
Show solution
- Since $f(x)$ is a polynomial, it is continuous on $[0, 2]$ and differentiable on $(0, 2)$.
- Check endpoint matches: $f(0) = 0$, $f(2) = 2(1)(0) = 0$. Thus, $f(0) = f(2) = 0$.
Cauchy Extensions and Taylor SeriesTopic 2
- Cauchy's Mean Value Theorem: If two functions $f(x)$ and $g(x)$ are continuous on $[a, b]$ and differentiable on $(a, b)$, with $g'(x) \neq 0$ for all $x \in (a, b)$, then there exists at least one $c \in (a, b)$ such that: \[ \frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)} \]
- Taylor's and Maclaurin's Series: If a function has derivatives of all orders throughout a neighborhood of $x=a$, it can be represented as a power series. When expanded around $a=0$, it is called a Maclaurin series: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \dots + \frac{f^{(n)}(0)}{n!}x^n + \dots \]
Expand $f(x) = \ln(1+x)$ around $x=0$ up to the third degree using Maclaurin series definitions.
Show solution
- $f(x) = \ln(1+x) \implies f(0) = \ln(1) = 0$
- $f'(x) = \frac{1}{1+x} \implies f'(0) = 1$
- $f''(x) = -\frac{1}{(1+x)^2} \implies f''(0) = -1$
- $f'''(x) = \frac{2}{(1+x)^3} \implies f'''(0) = 2$
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