Differentiation in Determinants: How to Differentiate a Determinant
Master the essential techniques for differentiating determinants with step-by-step methods and JEE-level practice problems.
Why Differentiation of Determinants is Important for JEE
Differentiation of determinants is a crucial topic that appears frequently in JEE Main and Advanced. Based on analysis of previous years' papers, this concept appears in 1-2 questions every year. Mastering these techniques will help you:
- Solve complex differentiation problems efficiently
- Handle parametric determinants with confidence
- Apply multiple differentiation methods strategically
- Save crucial time during the examination
Key Formula: Differentiation of Determinants
For a 2×2 determinant:
$\frac{d}{dx} \begin{vmatrix} f_1(x) & f_2(x) \\ g_1(x) & g_2(x) \end{vmatrix} = \begin{vmatrix} f_1'(x) & f_2'(x) \\ g_1(x) & g_2(x) \end{vmatrix} + \begin{vmatrix} f_1(x) & f_2(x) \\ g_1'(x) & g_2'(x) \end{vmatrix}$
For a 3×3 determinant:
$\frac{d}{dx} \begin{vmatrix} f_1 & f_2 & f_3 \\ g_1 & g_2 & g_3 \\ h_1 & h_2 & h_3 \end{vmatrix} = \begin{vmatrix} f_1' & f_2' & f_3' \\ g_1 & g_2 & g_3 \\ h_1 & h_2 & h_3 \end{vmatrix} + \begin{vmatrix} f_1 & f_2 & f_3 \\ g_1' & g_2' & g_3' \\ h_1 & h_2 & h_3 \end{vmatrix} + \begin{vmatrix} f_1 & f_2 & f_3 \\ g_1 & g_2 & g_3 \\ h_1' & h_2' & h_3' \end{vmatrix}$
Direct Differentiation Method
Differentiate each element systematically using the standard formula.
Example: Differentiate $D(x) = \begin{vmatrix} x & x^2 \\ x^3 & x^4 \end{vmatrix}$
Step 1: Apply the differentiation formula:
$\frac{dD}{dx} = \begin{vmatrix} 1 & 2x \\ x^3 & x^4 \end{vmatrix} + \begin{vmatrix} x & x^2 \\ 3x^2 & 4x^3 \end{vmatrix}$
Step 2: Calculate each determinant:
First determinant: $(1)(x^4) - (2x)(x^3) = x^4 - 2x^4 = -x^4$
Second determinant: $(x)(4x^3) - (x^2)(3x^2) = 4x^4 - 3x^4 = x^4$
Step 3: Add results: $\frac{dD}{dx} = -x^4 + x^4 = 0$
Final Answer: $\frac{dD}{dx} = 0$
Expansion Then Differentiation
First expand the determinant, then differentiate the resulting expression.
Example: Differentiate $D(x) = \begin{vmatrix} \sin x & \cos x \\ -\cos x & \sin x \end{vmatrix}$
Step 1: Expand the determinant:
$D(x) = (\sin x)(\sin x) - (\cos x)(-\cos x) = \sin^2 x + \cos^2 x = 1$
Step 2: Differentiate the result:
$\frac{dD}{dx} = \frac{d}{dx}(1) = 0$
Final Answer: $\frac{dD}{dx} = 0$
Parametric Determinants
Differentiate determinants where elements are functions of a parameter.
Example: If $D(t) = \begin{vmatrix} t & t^2 & t^3 \\ 1 & 2t & 3t^2 \\ 0 & 2 & 6t \end{vmatrix}$, find $\frac{dD}{dt}$
Step 1: Apply the 3×3 differentiation formula:
$\frac{dD}{dt} = \begin{vmatrix} 1 & 2t & 3t^2 \\ 1 & 2t & 3t^2 \\ 0 & 2 & 6t \end{vmatrix} + \begin{vmatrix} t & t^2 & t^3 \\ 0 & 2 & 6t \\ 0 & 2 & 6t \end{vmatrix} + \begin{vmatrix} t & t^2 & t^3 \\ 1 & 2t & 3t^2 \\ 0 & 0 & 6 \end{vmatrix}$
Step 2: Notice first two determinants have identical rows/columns:
First determinant = 0 (identical rows)
Second determinant = 0 (identical rows)
Step 3: Calculate third determinant:
$\begin{vmatrix} t & t^2 & t^3 \\ 1 & 2t & 3t^2 \\ 0 & 0 & 6 \end{vmatrix} = t \cdot \begin{vmatrix} 2t & 3t^2 \\ 0 & 6 \end{vmatrix} - t^2 \cdot \begin{vmatrix} 1 & 3t^2 \\ 0 & 6 \end{vmatrix} + t^3 \cdot \begin{vmatrix} 1 & 2t \\ 0 & 0 \end{vmatrix}$
$= t(12t) - t^2(6) + t^3(0) = 12t^2 - 6t^2 = 6t^2$
Final Answer: $\frac{dD}{dt} = 6t^2$
🚀 Quick Solving Strategies
When to Use Each Method:
- Direct Method: For simple determinants with differentiable elements
- Expansion Method: When expansion gives a simpler expression
- Parametric Method: For determinants with parameter-dependent elements
- Row/Column Operations: When determinants can be simplified first
Common Pitfalls to Avoid:
- Forgetting to differentiate all rows/columns systematically
- Missing the plus signs between derivative determinants
- Not checking if expansion would be simpler first
- Overlooking identical rows/columns that make determinants zero
Methods 4-6 Available in Full Version
Includes Higher Order Differentiation, Wronskian Determinants, and Applications in Differential Equations
JEE Advanced Level Problem
If $D(x) = \begin{vmatrix} f(x) & g(x) & h(x) \\ xf'(x) & xg'(x) & xh'(x) \\ x^2f''(x) & x^2g''(x) & x^2h''(x) \end{vmatrix}$, prove that $D'(x) = \frac{3}{x}D(x)$
Solution Approach:
Step 1: Apply the differentiation formula to $D(x)$
Step 2: Differentiate each row systematically
Step 3: Use product rule for elements with $x$ factors
Step 4: Show the relationship $D'(x) = \frac{3}{x}D(x)$
Key Insight: This demonstrates how derivative determinants can reveal functional relationships
📝 Quick Self-Test
Try these JEE-level problems to test your understanding:
1. Differentiate $D(x) = \begin{vmatrix} x^2 & \ln x \\ e^x & \sin x \end{vmatrix}$
2. Find $\frac{d}{dx} \begin{vmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{vmatrix}$
3. If $D(x) = \begin{vmatrix} 1 & x & x^2 \\ x & x^2 & 1 \\ x^2 & 1 & x \end{vmatrix}$, find $D'(1)$
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