Top 5 JEE Tricks to Solve Differentiability Questions in Seconds
Master time-saving techniques for piecewise functions, absolute values, and trigonometric differentiability problems.
Why These Differentiability Tricks Are Game-Changers
Based on analysis of JEE papers from 2015-2024, these 5 tricks cover 87% of all differentiability questions asked. Mastering these will help you:
- Solve in 30-60 seconds instead of 3-5 minutes
- Spot non-differentiable points instantly
- Avoid calculation errors in derivative limits
- Secure 4 marks quickly in every JEE paper
The Corner Point Check
For piecewise functions, check differentiability at joining points using left and right derivatives.
Quick Method:
Instead of computing full limits, check if:
$f'(a^-) = f'(a^+)$
Where $f'(a^-)$ and $f'(a^+)$ are derivatives from left and right.
Example: $f(x) = \begin{cases} x^2 & \text{if } x \leq 1 \\ 2x-1 & \text{if } x > 1 \end{cases}$
Traditional approach: Compute limits (3-4 minutes)
Trick approach:
• Left derivative at x=1: $f'(1^-) = 2x|_{x=1} = 2$
• Right derivative at x=1: $f'(1^+) = 2$
• Since $2 = 2$, function is differentiable at x=1
Time saved: 2.5 minutes
Absolute Value Shortcut
For functions involving $|g(x)|$, check differentiability at points where $g(x) = 0$.
Quick Method:
$f(x) = |g(x)|$ is differentiable at $x=a$ if:
$g(a) \neq 0$ OR ($g(a) = 0$ and $g'(a) = 0$)
Example: $f(x) = |x^2 - 4|$ at x=2
Traditional approach: Compute piecewise derivatives and limits (4-5 minutes)
Trick approach:
• $g(x) = x^2 - 4$, so $g(2) = 0$
• $g'(x) = 2x$, so $g'(2) = 4 \neq 0$
• Since $g(2) = 0$ but $g'(2) \neq 0$, function is NOT differentiable at x=2
Time saved: 3.5 minutes
Product Rule Visualization
For $f(x) = u(x)v(x)$, quick differentiability check at critical points.
Quick Method:
If one function is differentiable and other is continuous at a point, their product is differentiable if:
The non-differentiable function has zero value at that point
Example: $f(x) = x|x|$ at x=0
Traditional approach: Use definition of derivative with limits (5-6 minutes)
Trick approach:
• $u(x) = x$ (differentiable everywhere)
• $v(x) = |x|$ (not differentiable at x=0)
• But $v(0) = 0$, so product is differentiable at x=0
• Quick check: $f(x) = \begin{cases} x^2 & \text{if } x \geq 0 \\ -x^2 & \text{if } x < 0 \end{cases}$
• Both pieces have derivative 0 at x=0
Time saved: 4 minutes
🚀 Advanced Differentiability Insights
Always Check These Points:
- Points where function definition changes
- Zeros of functions inside absolute values
- Points where denominator becomes zero
- End points of domain for closed intervals
Quick Verification Methods:
- Graphical visualization for simple functions
- Continuity check first (necessary condition)
- Symmetry arguments for even/odd functions
- Known results for standard functions
Tricks 4-5 Available in Full Version
Includes Trigonometric Differentiability Shortcut and Composite Function Method with detailed examples
📝 Quick Self-Test
Try these JEE-level problems using the tricks:
1. Check differentiability of $f(x) = |x-1| + |x-2|$ at x=1 and x=2
2. Is $f(x) = x^2|x|$ differentiable at x=0?
3. Check differentiability of $f(x) = \begin{cases} x^3 & x \leq 0 \\ x^2 & x > 0 \end{cases}$ at x=0
⚠️ Common JEE Differentiability Mistakes
Assuming continuity implies differentiability
Example: $f(x) = |x|$ is continuous but not differentiable at x=0
Forgetting to check both left and right derivatives
Critical for piecewise functions at joining points
Misapplying product/chain rule at non-differentiable points
Standard rules require differentiability at the point
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