Problem 38 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f, g, h\) are differentiable functions of \(x\) and \(\Delta=\left|\begin{array}{ccc}f & g & h \\ (x f)^{\prime} & (x g)^{\prime} & (x h)^{\prime} \\ \left(x^{2} f\right)^{\prime \prime} & \left(x^{2} g\right)^{\prime \prime} & \left(x^{2} h\right)^{\prime \prime}\end{array}\right|\) then \(\Delta^{\prime}\) (the derivative of \(\Delta\) with respect to \(x\) ) is given by (A) \(\left|\begin{array}{ccc}f^{\prime} & g^{\prime} & h^{\prime} \\ f & g & h \\ \left(x^{3} f^{\prime \prime}\right)^{\prime} & \left(x^{3} g^{\prime \prime}\right)^{\prime} & \left(x^{3} h^{\prime \prime}\right)^{\prime}\end{array}\right|\) (B) \(\left|\begin{array}{ccc}f & g & h \\ f^{\prime} & g^{\prime} & h^{\prime} \\ \left(x^{2} f^{\prime \prime}\right)^{\prime} & \left(x^{2} g^{\prime \prime}\right)^{\prime} & \left(x^{2} h^{\prime \prime}\right)^{\prime}\end{array}\right|\) (C) \(\left|\begin{array}{ccc}f & g & h \\ f^{\prime} & g^{\prime} & h^{\prime} \\ \left(x^{3} f^{\prime \prime}\right)^{\prime} & \left(x^{3} g^{\prime \prime}\right)^{\prime} & \left(x^{3} h^{\prime \prime}\right)^{\prime}\end{array}\right|\) (D) None of these

Step-by-Step Solution

Answer

The correct answer is option (C).

Step 1: Define the Derivative of a Determinant

The derivative of a determinant \( \Delta \) with respect to \( x \) is given by differentiating each row, one at a time, while keeping the other rows the same. This is based on the property that the derivative of a determinant of functions is the sum of the determinants after differentiating each row individually.

Step 2: Differentiate the First Row

The first row of \( \Delta \) is \( [f, g, h] \). Upon differentiating, it becomes \( [f', g', h'] \). Keep the other rows as they are:\[\left| \begin{array}{ccc} f' & g' & h' \ (x f)^{\prime} & (x g)^{\prime} & (x h)^{\prime} \ \left(x^{2} f\right)^{\prime\prime} & \left(x^{2} g\right)^{\prime\prime} & \left(x^{2} h\right)^{\prime\prime} \end{array} \right|\]

Step 3: Differentiate the Second Row

The second row is \( [(x f)^{\prime}, (x g)^{\prime}, (x h)^{\prime}] \), which simplifies to \( [xf' + f, xg' + g, xh' + h] \). Differentiating, we get \( [xf'' + 2f', xg'' + 2g', xh'' + 2h'] \). Now the determinant is:\[\left| \begin{array}{ccc} f & g & h \ xf'' + 2f' & xg'' + 2g' & xh'' + 2h' \ \left(x^{2} f\right)^{\prime\prime} & \left(x^{2} g\right)^{\prime\prime} & \left(x^{2} h\right)^{\prime\prime} \end{array} \right| \]

Step 4: Differentiate the Third Row

The third row is \( \left[\left(x^{2} f\right)^{\prime\prime}, \left(x^{2} g\right)^{\prime\prime}, \left(x^{2} h\right)^{\prime\prime}\right] \), which simplifies to \( [x^2f'' + 4xf' + 2f, x^2g'' + 4xg' + 2g, x^2h'' + 4xh' + 2h] \). Differentiating gives us \((x^3f''')^{\prime}, (x^3g''')^{\prime}, (x^3h''')^{\prime}\). Insert into the determinant:\[\left| \begin{array}{ccc} f & g & h \ (x f)^{\prime} & (x g)^{\prime} & (x h)^{\prime} \ (x^{3}f''')^{\prime} & (x^{3}g''')^{\prime} & (x^{3}h''')^{\prime} \end{array} \right| \]

Step 5: Combine the Terms

Now, we combine the determinants calculated in steps 2, 3, and 4 using the property of the derivative of a determinant:\[ \Delta' = \left| \begin{array}{ccc} f' & g' & h' \ (x f)^{\prime} & (x g)^{\prime} & (x h)^{\prime} \ \left(x^{2} f\right)^{\prime\prime} & \left(x^{2} g\right)^{\prime\prime} & \left(x^{2} h\right)^{\prime\prime} \end{array} \right|+ \left| \begin{array}{ccc} f & g & h \ f' & g' & h' \ \left(x^{3} f^{\prime\prime}\right)^{\prime} & \left(x^{3} g^{\prime\prime}\right)^{\prime} & \left(x^{3} h^{\prime\prime}\right)^{\prime} \end{array} \right| \] Compare it with the options to find the correct expression.

Step 6: Match with Multiple Choice

On comparing the expanded form with the options provided, we see that option C \(\left|\begin{array}{ccc}f & g & h \ f^{\prime} & g^{\prime} & h^{\prime} \ (x^{3} f^{\prime\prime})^{\prime} & (x^{3} g^{\prime\prime})^{\prime} & (x^{3} h^{\prime\prime})^{\prime} \end{array}\right|\) agrees with the part of the expanded expression we derived, and thus option C is the solution.

Key Concepts

Differentiable Functions

A function is said to be differentiable if it has a derivative at every point in its domain. This means that the function's graph has a tangent line at every point, and no sharp corners or discontinuities interrupt the curve. In simpler terms, a differentiable function is one you can smoothly sketch without lifting your pen from the paper.
A few key points about differentiable functions include:

• If a function is differentiable, it is also continuous. However, continuity does not guarantee differentiability.
• Polynomials, sine, cosine, and exponential functions are some classic examples of differentiable functions.
• We often denote the derivative of a function, say \( f(x) \), by \( f'(x) \) or \( \frac{df}{dx} \).

Differentiable functions are foundational in calculus and pave the way for more complex transformations, such as finding the derivative of a determinant like in the current exercise.

Second Derivative

The second derivative of a function is simply the derivative of the derivative. In essence, it tells us how the rate of change itself is changing. If \( f(x) \) represents our original function, then \( f''(x) \) represents the second derivative.
Understanding the second derivative is crucial for several reasons:

• The second derivative helps identify concavity of the function. If \( f''(x) > 0 \), the function is concave up, and if \( f''(x) < 0 \), it is concave down.
• It helps us determine potential points of inflection where the curvature of the function changes.
• In applied mathematics, second derivatives can signify acceleration if the original function represents velocity.

In our task, the second derivatives \( (x^2 f)'' \), \( (x^2 g)'' \), and \( (x^2 h)'' \) form part of the matrix whose determinant we are differentiating.

Matrix Differentiation

Matrix differentiation is an extension of traditional calculus applied to matrices, offering a powerful tool for dealing with systems of equations or multiple variable functions. When dealing with determinants of matrices, each element can depend on some variable, like \( x \).
There's a notable concept when differentiating determinants:

• We differentiate each element of the determinant matrix row by row while keeping the others constant, reflecting a property of partial derivatives.
• After differentiating, the new rows are combined into new determinants that explain changes in specific dimensions related to the original function.
• This process becomes essential when computing derivatives in multivariable calculus or linear algebra applications.

In this exercise, differentiating the matrix systematically helped us derive the derivative of the determinant, ultimately leading us to the correct option among the multiple choices. Understanding this concept enhances your ability to tackle complex calculus problems involving matrices.