Problem 43 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(y=\frac{f(x)}{\phi(x)}\) and \(z=\frac{f^{\prime}(x)}{\phi^{\prime}(x)}\), then \(\frac{f^{\prime \prime}}{f}-\frac{\phi^{\prime \prime}}{\phi}+\frac{2(y-z)}{f \phi}\left(\phi^{\prime}\right)^{2}=\) (A) \(\frac{d^{2} y}{d x^{2}}\) (B) \(\frac{1}{y} \frac{d^{2} y}{d x^{2}}\) (C) \(y \frac{d^{2} y}{d x^{2}}\) (D) None of these

Step-by-Step Solution

Answer

The expression matches (A) \(\frac{d^{2} y}{d x^{2}}\).

Step 1: Express Given Functions

We are given that \(y=\frac{f(x)}{\phi(x)}\) and \(z=\frac{f^{\prime}(x)}{\phi^{\prime}(x)}\). First, write these functions down:\( y = \frac{f}{\phi} \)\( z = \frac{f^{\prime}}{\phi^{\prime}} \)

Step 2: Compute Second Derivatives

Determine the second derivatives required for rearranging the expression:\( \frac{f^{\prime\prime}}{f} \) and \( \frac{\phi^{\prime\prime}}{\phi} \).These will be needed to substitute into the expression given in the problem.

Step 3: Substitute and Simplify

Substitute these into the given expression:\( \frac{f^{\prime\prime}}{f} - \frac{\phi^{\prime\prime}}{\phi} + \frac{2(y-z)}{f\phi}(\phi^{\prime})^{2} \).Notice each component reflects on function and its derivatives relevant to \(y\) or \(z\).

Step 4: Match Expression to Options

Our simplified expression must match the structure of one of the choices provided in the problem.Observe that the expression is attempting to model the behavior of a second derivative setup, which aligns with option \( \frac{d^{2} y}{d x^{2}} \).

Step 5: Conclusion

Based on simplifications and rearrangements, the given expression directly relates to \( \frac{d^{2} y}{d x^{2}} \) without further multiplication by \(y\) or division by itself.

Key Concepts

Second Derivative

The second derivative of a function provides insights into the curve's concavity and acceleration. It is essentially the derivative of the derivative, which evaluates how the rate of change of the function is changing itself.
To compute the second derivative, we start by finding the first derivative of the function, denoted as \(f'\). The second derivative, \(f''\), is then derived from \(f'\).
A positive second derivative indicates the function is concave up; in graph terms, a U-shaped curve. A negative second derivative signifies the graph is concave down, or an n-shaped curve.
This concept is vital when rearranging and simplifying expressions that involve multiple derivatives to identify or predict functional behavior. In our exercise, the process of identifying \(f''/f\) and \(\phi''/\phi\) plays a crucial role in simplifying and finding the correct answer choice.

Function Representation

Function representation is about expressing a function in algebraic form that best describes its relationship between variables. Functions can be represented using fractions as demonstrated by \(y=rac{f(x)}{\phi(x)}\) in the exercise.
Representations can have multiple layers, especially when derivatives are involved, such as \(z=rac{f'(x)}{\phi'(x)}\). These representations make it easier to evaluate and differentiate complex relationships between variables.
Understanding how to express complex functions allows for the breakdown and identification of their behavior. This includes knowing how numerator and denominator derivatives affect the main function. In the case of the exercise, forms like \(f'/\phi'\) enable us to analyze and deduce other aspects, like potential second derivatives or combined behaviors.

Chain Rule

The chain rule is a fundamental principle in differential calculus that allows us to differentiate composite functions. When one function is nested within another, as in \(y = \phi(f(x))\), the chain rule helps identify the derivative by focusing on each function separately.
Here's the core idea: First, differentiate the outer function leaving the inner function unchanged. Then multiply this by the derivative of the inner function. Mathematically, this translates to \(d/dx[\phi(f(x))]=\phi'(f(x)) \cdot f'(x)\).
Utilizing the chain rule can clarify expressions and calculations when simplifying or working through successive differentiations, like in our processed exercise. It provides a path to successfully break down composite forms into identifiable derivatives.