Differentiability in Intervals | JEE Main Calculus Crash Course | Mathematicon

Differentiability in Intervals: Your JEE Main Ticket

Hello JEE aspirants! Differentiability is a cornerstone of calculus, and mastering it within intervals is crucial for acing those JEE Main questions. We're going to break down the concepts, formulas, and tricks to make you a differentiability ninja. Let's begin!

1. Differentiability on Open Intervals (a, b)

Imagine a smooth, continuous curve. Differentiability on an open interval is like saying you can zoom in infinitely on any point between $a$ and $b$ and the curve will always look like a straight line. No sharp corners, no breaks!

Concept: A function $f(x)$ is differentiable on the open interval $(a, b)$ if the derivative $f'(x)$ exists for every $x$ within that interval. This means that at each point, you can find the slope of the tangent.

f \text{ is differentiable on } (a,b) \text{ if } f'(x) \text{ exists for all } x \in (a,b)

Intuition: Think of $f'(x)$ as the instantaneous rate of change. If this rate exists at every point in $(a, b)$ , you're golden! For instance, $f(x) = x^2$ is differentiable on $(-1, 1)$ because $f'(x) = 2x$ exists for all $x$ in that interval.

2. Differentiability on Closed Intervals [a, b]: Handling the Endpoints

Closed intervals include the endpoints, and that's where things get slightly trickier. You need to ensure the function is well-behaved at the boundaries.

Concept: A function $f(x)$ is differentiable on the closed interval $[a, b]$ if:

$f'(x)$ exists for all $x$ in the open interval $(a, b)$ .
The right-hand derivative (RHD) exists at $x = a$ .
The left-hand derivative (LHD) exists at $x = b$ .

f \text{ is differentiable on } [a,b] \text{ if } f'(x) \text{ exists on } (a,b), \text{ RHD exists at } a, \text{ LHD exists at } b

RHD and LHD:

Right-Hand Derivative (RHD) at $a$ : $f'(a^+) = \lim_{h \rightarrow 0^+} \frac{f(a+h) - f(a)}{h}$
Left-Hand Derivative (LHD) at $b$ : $f'(b^-) = \lim_{h \rightarrow 0^-} \frac{f(b+h) - f(b)}{h}$

Intuition: Imagine you're standing at endpoint $a$ . You can only look to the right (towards the interval). If the slope of the tangent exists as you approach $a$ from the right, RHD exists. Similarly, at endpoint $b$ , you look left.

Example: Consider $f(x) = \sqrt{x}$ on $[0, 1]$ . $f'(x) = \frac{1}{2\sqrt{x}}$ exists for $(0, 1)$ . At $x = 0$ , the RHD is $\lim_{h \rightarrow 0^+} \frac{\sqrt{0+h} - \sqrt{0}}{h} = \lim_{h \rightarrow 0^+} \frac{1}{\sqrt{h}} = \infty$ . Since RHD doesn't exist as a finite number, $f(x)$ is not differentiable on $[0, 1]$ .

3. Corner Points and Cusps: The Differentiability Killers

Corner points and cusps are points where a function is continuous but not differentiable. They are like sharp turns in the road – you can't define a unique tangent at these points.

Corner Point: At a corner point, the LHD and RHD exist, but they are not equal.

Cusp: At a cusp, the LHD and RHD both exist, but have opposite signs and tend to infinity.

Example: $f(x) = |x|$ has a corner point at $x = 0$ . LHD at $0$ is $-1$ , and RHD at $0$ is $1$ . Since LHD ≠ RHD, $f(x)$ is not differentiable at $x = 0$ .

4. Vertical Tangents: Slope Goes Infinite!

A vertical tangent occurs when the derivative approaches infinity (or negative infinity) at a certain point. While the function might be continuous, it's not differentiable at that point.

Example: $f(x) = x^{1/3}$ has a vertical tangent at $x = 0$ . $f'(x) = \frac{1}{3x^{2/3}}$ , which approaches infinity as $x$ approaches $0$ .

5. Continuous but Not Differentiable: The Subtle Trap

Continuity is a necessary but not sufficient condition for differentiability. A function can be continuous at a point, but still not differentiable there.

Key Reason: Sharp changes in direction, like corners, cusps, or vertical tangents, break differentiability even if the function is continuous.

Example: As we saw, $f(x) = |x|$ is continuous at $x = 0$ , but not differentiable.

Tip: When checking for differentiability, ALWAYS check for continuity first. If a function is discontinuous, it's automatically not differentiable. But remember, continuity doesn't guarantee differentiability!

Common Mistake: Assuming that a function is differentiable just because you can draw its graph without lifting your pen. Always check LHD and RHD at suspicious points!

JEE-Specific Tricks

Piecewise Functions: These are favorites for JEE. Carefully check the differentiability at the points where the function definition changes.
Implicit Differentiation: Understand how to find $dy/dx$ when $y$ is not explicitly defined in terms of $x$ . Always simplify your final answer.
Higher-Order Derivatives: Questions involving second derivatives and their applications (e.g., concavity) are common.

Example (JEE-style): Let $f(x) = \begin{cases} ax^2 + b & \text{if } x < 1 \\ x + 1 & \text{if } x \geq 1 \end{cases}$ . Find the relationship between $a$ and $b$ such that $f(x)$ is differentiable at $x = 1$ .

Solution:

Continuity: $a(1)^2 + b = 1 + 1 \Rightarrow a + b = 2$ .
Differentiability: $f'(x) = \begin{cases} 2ax & \text{if } x < 1 \\ 1 & \text{if } x \geq 1 \end{cases}$ . $2a(1) = 1 \Rightarrow a = \frac{1}{2}$ . Therefore, $b = 2 - \frac{1}{2} = \frac{3}{2}$ .

Therefore, for $f(x)$ to be differentiable at $x=1$ , $a = \frac{1}{2}$ and $b = \frac{3}{2}$ .

Tip: For piecewise functions, equate the function values and the derivatives at the transition points to ensure differentiability.

Common Mistake: Forgetting to check both continuity AND differentiability for piecewise functions. Continuity is a prerequisite!

Alright, future IITians! You're now armed with the knowledge to tackle differentiability in intervals. Keep practicing, stay sharp, and ace those JEE Main exams!