Differentiability and Derivatives up to Second Order
Introduction to Differentiability
Differentiability is a fundamental concept in calculus that tells us whether a function can be "smoothly" differentiated at a particular point. Intuitively, a function is differentiable at a point if its graph has a well-defined, non-vertical tangent line at that point. This means the graph doesn't have any sharp corners, cusps, or breaks.
Mathematically, a function $f(x)$ is said to be differentiable at a point $x = c$ if the limit $$ \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} $$ exists. This limit is called the derivative of $f(x)$ at $x = c$, denoted by $f'(c)$ or $\frac{dy}{dx}\Big|_{x=c}$.
For the derivative to exist, the left-hand derivative and the right-hand derivative must be equal.
- Right-hand derivative: $f'(c^+) = \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h}$
- Left-hand derivative: $f'(c^-) = \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h}$
A function $f(x)$ is differentiable at $x=c$ if and only if $f'(c^+) = f'(c^-)$.
Conditions for Differentiability
For a function $f(x)$ to be differentiable at a point $x = c$, two conditions must be met:
- The function $f(x)$ must be continuous at $x = c$.
- The left-hand derivative must equal the right-hand derivative at $x = c$.
If a function is differentiable at a point, it is necessarily continuous at that point. However, the converse is not always true; a function can be continuous at a point but not differentiable there.
Examples of Non-Differentiable Points
Consider the function $f(x) = |x|$ at $x=0$.
$f(0) = |0| = 0$.
Right-hand derivative at $x=0$: $$ \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{|h| - 0}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1 $$
Left-hand derivative at $x=0$: $$ \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{|h| - 0}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1 $$
Since $f'(0^+) \neq f'(0^-)$ (1 $\neq$ -1), the function $f(x) = |x|$ is not differentiable at $x=0$. This corresponds to the sharp corner in the graph of $y = |x|$ at the origin.
Rules of Differentiation
To find derivatives of complex functions, we use several standard rules. These rules are derived from the definition of the derivative.
1. Power Rule
If $f(x) = x^n$, then $f'(x) = nx^{n-1}$.
Example: If $f(x) = x^3$, then $f'(x) = 3x^{3-1} = 3x^2$.
Example: If $f(x) = \sqrt{x} = x^{1/2}$, then $f'(x) = \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.
2. Constant Multiple Rule
If $f(x) = k \cdot g(x)$, where $k$ is a constant, then $f'(x) = k \cdot g'(x)$.
Example: If $f(x) = 5x^4$, then $f'(x) = 5 \cdot (4x^{4-1}) = 5 \cdot 4x^3 = 20x^3$.
3. Sum and Difference Rules
If $f(x) = g(x) \pm h(x)$, then $f'(x) = g'(x) \pm h'(x)$.
Example: If $f(x) = x^2 + 3x$, then $f'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(3x) = 2x + 3$.
4. Product Rule
If $f(x) = u(x) \cdot v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.
Example: If $f(x) = x^2 \sin(x)$, let $u(x) = x^2$ and $v(x) = \sin(x)$. Then $u'(x) = 2x$ and $v'(x) = \cos(x)$. So, $f'(x) = (2x)\sin(x) + x^2(\cos(x)) = 2x\sin(x) + x^2\cos(x)$.
5. Quotient Rule
If $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.
Example: If $f(x) = \frac{x^2}{e^x}$, let $u(x) = x^2$ and $v(x) = e^x$. Then $u'(x) = 2x$ and $v'(x) = e^x$. So, $f'(x) = \frac{(2x)(e^x) - (x^2)(e^x)}{(e^x)^2} = \frac{e^x(2x - x^2)}{e^{2x}} = \frac{2x - x^2}{e^x}$.
6. Chain Rule
If $f(x) = g(h(x))$, then $f'(x) = g'(h(x)) \cdot h'(x)$. This rule is used for composite functions.
Example: If $f(x) = \sin(x^2)$, let $g(u) = \sin(u)$ and $h(x) = x^2$. Then $g'(u) = \cos(u)$ and $h'(x) = 2x$. So, $f'(x) = g'(h(x)) \cdot h'(x) = \cos(x^2) \cdot (2x) = 2x\cos(x^2)$.
Example: If $f(x) = (x^3 + 1)^5$, let $g(u) = u^5$ and $h(x) = x^3 + 1$. Then $g'(u) = 5u^4$ and $h'(x) = 3x^2$. So, $f'(x) = g'(h(x)) \cdot h'(x) = 5(x^3 + 1)^4 \cdot (3x^2) = 15x^2(x^3 + 1)^4$.
Derivatives of Standard Functions
It's essential to memorize the derivatives of common functions.
| Function $f(x)$ | Derivative $f'(x)$ |
|---|---|
| $c$ (constant) | $0$ |
| $x^n$ | $nx^{n-1}$ |
| $e^x$ | $e^x$ |
| $a^x$ ($a>0, a \neq 1$) | $a^x \ln a$ |
| $\ln x$ ($x>0$) | $\frac{1}{x}$ |
| $\log_a x$ ($x>0, a>0, a \neq 1$) | $\frac{1}{x \ln a}$ |
| $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ |
| $\tan x$ | $\sec^2 x$ |
| $\cot x$ | $-\csc^2 x$ |
| $\sec x$ | $\sec x \tan x$ |
| $\csc x$ | $-\csc x \cot x$ |
| $\sin^{-1} x$ (domain: $[-1, 1]$) | $\frac{1}{\sqrt{1-x^2}}$ (domain: $(-1, 1)$) |
| $\cos^{-1} x$ (domain: $[-1, 1]$) | $\frac{-1}{\sqrt{1-x^2}}$ (domain: $(-1, 1)$) |
| $\tan^{-1} x$ | $\frac{1}{1+x^2}$ |
| $\cot^{-1} x$ | $\frac{-1}{1+x^2}$ |
| $\sec^{-1} x$ (domain: $(-\infty, -1] \cup [1, \infty)$) | $\frac{1}{|x|\sqrt{x^2-1}}$ (domain: $(-\infty, -1) \cup (1, \infty)$) |
| $\csc^{-1} x$ (domain: $(-\infty, -1] \cup [1, \infty)$) | $\frac{-1}{|x|\sqrt{x^2-1}}$ (domain: $(-\infty, -1) \cup (1, \infty)$) |
Implicit Differentiation
Implicit differentiation is a technique used to find the derivative of an equation where $y$ is not explicitly defined as a function of $x$. We differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ and using the chain rule.
Steps:
- Differentiate both sides of the equation with respect to $x$. Remember to apply the chain rule whenever you differentiate a term involving $y$. For example, $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$.
- Isolate $\frac{dy}{dx}$ by algebraic manipulation.
Example: Find $\frac{dy}{dx}$ for the equation $x^2 + y^2 = 25$.
Differentiate both sides with respect to $x$: $$ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25) $$ $$ 2x + 2y \frac{dy}{dx} = 0 $$
Now, solve for $\frac{dy}{dx}$: $$ 2y \frac{dy}{dx} = -2x $$ $$ \frac{dy}{dx} = \frac{-2x}{2y} = -\frac{x}{y} $$
Example: Find $\frac{dy}{dx}$ for $x^3 + y^3 = 6xy$.
Differentiate both sides with respect to $x$: $$ \frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(6xy) $$ $$ 3x^2 + 3y^2 \frac{dy}{dx} = 6 \left( 1 \cdot y + x \cdot \frac{dy}{dx} \right) \quad \text{(using product rule on the right side)} $$ $$ 3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x \frac{dy}{dx} $$
Group terms with $\frac{dy}{dx}$ on one side: $$ 3y^2 \frac{dy}{dx} - 6x \frac{dy}{dx} = 6y - 3x^2 $$ $$ \frac{dy}{dx} (3y^2 - 6x) = 6y - 3x^2 $$
Solve for $\frac{dy}{dx}$: $$ \frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{3(2y - x^2)}{3(y^2 - 2x)} = \frac{2y - x^2}{y^2 - 2x} $$
Derivatives of Higher Order
The derivative of a function $f(x)$ is itself a function, denoted by $f'(x)$ or $\frac{dy}{dx}$. We can differentiate this derivative to obtain the second derivative, and so on.
The first derivative, $f'(x)$, represents the rate of change of the function.
The second derivative, $f''(x)$ or $\frac{d^2y}{dx^2}$, represents the rate of change of the first derivative. It provides information about the concavity of the function's graph.
The third derivative, $f'''(x)$ or $\frac{d^3y}{dx^3}$, and so on, are called higher-order derivatives.
Calculating Second Order Derivatives
To find the second derivative, we simply differentiate the first derivative.
Example: Find the second derivative of $f(x) = x^4 + 3x^2 - 5$.
First derivative: $$ f'(x) = \frac{d}{dx}(x^4 + 3x^2 - 5) = 4x^3 + 6x $$
Second derivative (differentiate $f'(x)$): $$ f''(x) = \frac{d}{dx}(4x^3 + 6x) = 12x^2 + 6 $$
Example: Find the second derivative of $y = e^{3x}$.
First derivative (using chain rule): $$ \frac{dy}{dx} = \frac{d}{dx}(e^{3x}) = e^{3x} \cdot \frac{d}{dx}(3x) = e^{3x} \cdot 3 = 3e^{3x} $$
Second derivative (differentiate $\frac{dy}{dx}$): $$ \frac{d^2y}{dx^2} = \frac{d}{dx}(3e^{3x}) = 3 \cdot \frac{d}{dx}(e^{3x}) = 3 \cdot (e^{3x} \cdot 3) = 9e^{3x} $$
Second Derivative Test for Concavity
The second derivative plays a crucial role in determining the concavity of a function and finding local extrema.
- If $f''(x) > 0$ on an interval, the function $f(x)$ is concave up (shaped like a cup) on that interval.
- If $f''(x) < 0$ on an interval, the function $f(x)$ is concave down (shaped like a frown) on that interval.
- If $f''(c) = 0$ or $f''(c)$ is undefined, $x=c$ is a potential inflection point, where the concavity might change.
Second Derivative Test for Local Extrema
Suppose $f'(c) = 0$. This means $x=c$ is a critical point.
- If $f''(c) > 0$, then $f(x)$ has a local minimum at $x=c$.
- If $f''(c) < 0$, then $f(x)$ has a local maximum at $x=c$.
- If $f''(c) = 0$, the test is inconclusive, and we must use the first derivative test.
This test is particularly useful because it avoids the need to check the sign of the first derivative on either side of the critical point.
Applications of Derivatives
Derivatives have wide-ranging applications in various fields:
- Physics: Velocity is the first derivative of position with respect to time, and acceleration is the second derivative.
- Economics: Marginal cost and marginal revenue are derivatives of total cost and total revenue functions, respectively.
- Optimization: Finding maximum or minimum values of quantities (e.g., maximizing profit, minimizing material usage).
- Related Rates: Problems where the rates of change of different quantities are related.
- Curve Sketching: Using first and second derivatives to analyze the behavior of a function (increasing/decreasing intervals, concavity, local extrema).
L'Hôpital's Rule (Brief Mention for Context)
While not strictly within the scope of "derivatives up to second order," L'Hôpital's Rule is a crucial application of derivatives for evaluating indeterminate forms of limits like $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
If $\lim_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.