Differentiability

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Definition: $f$ is differentiable at $x = a$ if the derivative exists: $$f’(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$$

One-Sided Derivatives:

• Left-hand derivative: $f’-(a) = \lim{h \to 0^-} \frac{f(a+h) - f(a)}{h}$
• Right-hand derivative: $f’+(a) = \lim{h \to 0^+} \frac{f(a+h) - f(a)}{h}$

$f$ is differentiable at $a$ iff $f’-(a) = f’+(a)$ (finite).

Key Theorem: If $f$ is differentiable at $a$, then $f$ is continuous at $a$. (Converse is FALSE: $f(x) = |x|$ is continuous at $0$ but not differentiable there.)

Standard Non-Differentiable Cases:

1. Corner: Left and right derivatives exist but are different (e.g., $|x|$ at $0$)
2. Cusp: One-sided derivatives are infinite (e.g., $|x|^{1/3}$ at $0$)
3. Vertical tangent: $f’(a) = \infty$ (e.g., $x^{1/3}$ at $0$)

⚡ JEE Tip: When checking differentiability of piecewise functions, always compute both one-sided derivatives and check if they’re equal.

⚡ Common Mistake: Thinking that continuity implies differentiability. Remember: every differentiable function is continuous, but many continuous functions are not differentiable (corners, cusps, oscillations).

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Differentiability and Continuity:

• Theorem: $f$ differentiable at $a$ $\Rightarrow$ $f$ continuous at $a$.
• Proof: $\lim_{x \to a} [f(x) - f(a)] = \lim_{x \to a} \frac{f(x)-f(a)}{x-a} \cdot (x-a) = f’(a) \cdot 0 = 0$.

When NOT differentiable:

1. Corner point: $f(x) = |x|$ at $0$: left derivative $= -1$, right derivative $= +1$, not equal.

2. Cusp (sharp point): $f(x) = \sqrt{|x|} = |x|^{1/2}$ at $0$: left derivative $= -\infty$, right derivative $= +\infty$.

3. Oscillatory discontinuity: $f(x) = x \sin(1/x)$ for $x \neq 0$, $f(0) = 0$ is continuous but not differentiable at $0$. Also $f(x) = \sin(1/x^2)$ oscillates infinitely as $x \to 0$.

Differentiability on an Interval:

• $f$ is differentiable on $(a,b)$ if differentiable at every point in $(a,b)$
• $f$ is differentiable on $[a,b]$ if differentiable on $(a,b)$ and right differentiable at $a$ and left differentiable at $b$

Rules of Differentiation:

• Product: $(fg)’ = f’g + fg’$
• Quotient: $\left(\frac{f}{g}\right)’ = \frac{f’g - fg’}{g^2}$
• Chain: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Worked Examples:

Example 1: Check differentiability of $f(x) = |x^2 - 4|$ at $x = 2$.

At $x = 2$, $f(2) = |4-4| = 0$. For $x$ near 2: $f(x) = |(x-2)(x+2)|$. Since $x+2 \approx 4 > 0$ near $x=2$, sign is determined by $x-2$. So for $x < 2$, $f(x) = -(x-2)(x+2) = -(x^2-4) = 4-x^2$. For $x > 2$, $f(x) = (x-2)(x+2) = x^2-4$. $f’-(2) = \lim{h \to 0^-} \frac{f(2+h) - f(2)}{h} = \lim_{h \to 0^-} \frac{4-(2+h)^2}{h} = \lim_{h \to 0^-} \frac{4-4-4h-h^2}{h} = \lim_{h \to 0^-} (-4 - h) = -4$. $f’+(2) = \lim{h \to 0^+} \frac{(2+h)^2-4}{h} = \lim_{h \to 0^+} \frac{4+4h+h^2-4}{h} = \lim_{h \to 0^+} (4 + h) = 4$. Since $-4 \neq 4$, not differentiable at $x=2$.

Example 2 (JEE 2021): Find $k$ such that $f(x) = \begin{cases} kx^2 & x \leq 1 \ x + k & x > 1 \end{cases}$ is differentiable at $x=1$.

First, for differentiability, must be continuous: $f(1) = k(1)^2 = k$. $\lim_{x \to 1^-} f(x) = k$. $\lim_{x \to 1^+} f(x) = 1 + k$. For continuity at 1: $k = 1 + k$ → $0 = 1$, contradiction. So no value of $k$ makes $f$ continuous at $x=1$, hence not differentiable. Wait, the problem might have intended a form where continuity is possible.

Actually, maybe we want $k$ in the second piece? Let me recompute. $f(1) = k \cdot 1^2 = k$ from left. $\lim_{x \to 1^+} f(x) = 1 + k$. For continuity: $k = 1 + k$ → $0 = 1$. Impossible.

So answer is: no such $k$ exists.

Example 3: Is $f(x) = x^2 |x|$ differentiable everywhere?

$f(x) = x^2 \cdot |x| = x^2 \cdot \text{sgn}(x) \cdot x = x^3$ for $x > 0$? Actually $x^2 |x| = x^2 \cdot |x| = x^2 \cdot x$ for $x \geq 0$ and $x^2 \cdot (-x) = -x^3$ for $x < 0$. So $f(x) = \begin{cases} -x^3 & x < 0 \ x^3 & x \geq 0 \end{cases} = x^3$ (since $|x| = x$ for $x \geq 0$ and $-x$ for $x < 0$, but $x^2 (-x) = -x^3$ and $x^2 (x) = x^3$).

Wait: $x^2 |x| = x^2 \cdot |x|$. If $x \geq 0$: $|x| = x$, so $f(x) = x^3$. If $x < 0$: $|x| = -x$, so $f(x) = -x^3$.

At $x = 0$: $f(0) = 0$. Left derivative: $\lim_{h \to 0^-} \frac{-h^3 - 0}{h} = \lim_{h \to 0^-} \frac{-h^3}{h} = \lim_{h \to 0^-} (-h^2) = 0$. Right derivative: $\lim_{h \to 0^+} \frac{h^3}{h} = \lim_{h \to 0^+} h^2 = 0$. Both equal 0, so differentiable at 0 (and everywhere else since $x^3$ is differentiable). So $x^2|x|$ IS differentiable everywhere.

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🔴 Extended — Deep Study (3mo+)

Comprehensive theory for serious JEE Advanced preparation.

Derivative as Linear Approximation:

If $f$ is differentiable at $a$, then $f(x) = f(a) + f’(a)(x-a) + o(|x-a|)$ as $x \to a$. The linear function $L(x) = f(a) + f’(a)(x-a)$ is the best linear approximation to $f$ near $a$.

Differential: $df = f’(a) dx$ represents the differential of $f$. For small $\Delta x = dx$, $\Delta f \approx df$.

Implicit Differentiation:

If $F(x,y) = 0$ defines $y$ as a function of $x$, then: $$\frac{dy}{dx} = -\frac{\partial F/\partial x}{\partial F/\partial y}$$ provided denominator $\neq 0$.

Higher Order Derivatives:

$f^{(n)}(x) = \frac{d^n}{dx^n} f(x)$. For implicit functions, differentiate repeatedly.

Advanced Problems:

Problem (JEE Advanced 2023): If $f$ is differentiable at $x = 1$ and $\lim_{x \to 1} \frac{f(x) - f(1)}{x-1} = 4$, find $f’(1)$.

By definition, $f’(1) = \lim_{x \to 1} \frac{f(x) - f(1)}{x-1}$. Given this limit is 4, $f’(1) = 4$.

Problem 2: Let $f(x) = \begin{cases} x^2 \sin(1/x) & x \neq 0 \ 0 & x = 0 \end{cases}$. Is $f$ differentiable at $0$?

$f’(0) = \lim_{h \to 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h \to 0} h \sin(1/h)$. Since $|h \sin(1/h)| \leq |h| \to 0$, we have $f’(0) = 0$. So $f$ is differentiable at 0.

Now check $f’(x)$ for $x \neq 0$: $f’(x) = 2x \sin(1/x) + x^2 \cos(1/x) \cdot (-1/x^2) = 2x \sin(1/x) - \cos(1/x)$. At $x = 0$: $\lim_{x \to 0} f’(x) = \lim_{x \to 0} [2x \sin(1/x) - \cos(1/x)]$. The first term $\to 0$, but $\cos(1/x)$ oscillates between $-1$ and $1$. So $\lim_{x \to 0} f’(x)$ does not exist. Thus $f’$ is not continuous at $0$.

This shows that even if $f$ is differentiable everywhere, $f’$ need not be continuous.

JEE Advanced Patterns (2018–2024):

• Differentiability of piecewise functions with parameters is very common
• One-sided derivative calculations appear frequently
• Implicit differentiation and its higher order variants are tested
• Chain rule applications for complex compositions are common
• L’Hôpital’s rule and derivative in limit problems appeared in 2020, 2022

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