Problem 118 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\frac{d^{2} x}{d y^{2}}\) is equal to [2011] (A) \(-\left(\frac{d^{2} y}{d x^{2}}\right)^{-1}\left(\frac{d y}{d x}\right)^{-3}\) (B) \(\left(\frac{d^{2} y}{d x^{2}}\right)\left(\frac{d y}{d x}\right)^{-2}\) (C) \(-\left(\frac{d^{2} y}{d x^{2}}\right)\left(\frac{d y}{d x}\right)^{-3}\) (D) \(\left(\frac{d^{2} y}{d x^{2}}\right)^{-1}\)

Step-by-Step Solution

Answer

The answer is (C) \(-\left(\frac{d^{2} y}{d x^{2}}\right)\left(\frac{d y}{d x}\right)^{-3}\).

Step 1: Understand the Goal

We need to find an equivalent expression for \(\frac{d^{2} x}{d y^{2}}\). We are given expressions in terms of derivatives with \(d y\) and \(d x\) as variables.

Step 2: Use the Relation of Derivatives

Remember the chain rule, which states that the reciprocal of the derivative of a function is the derivative of the inverse function. So, \(\frac{d x}{d y} = \left(\frac{d y}{d x}\right)^{-1}\).

Step 3: Differentiate the Inverse Function

Differentiate \(\frac{d x}{d y}\) with respect to \(y\) using the chain rule:\[\frac{d^2 x}{d y^2} = \frac{d}{d y}\left(\frac{d x}{d y}\right)\]Substitute \(\frac{d x}{d y} = \left(\frac{d y}{d x}\right)^{-1}\) into this expression.

Step 4: Apply the Chain Rule

Applying the chain rule, we differentiate:\[\frac{d}{d y}\left(\left(\frac{d y}{d x}\right)^{-1}\right) = \frac{d}{d x}\left(\left(\frac{d y}{d x}\right)^{-1}\right) \cdot \frac{d x}{d y}\]The derivative of \(\left(\frac{d y}{d x}\right)^{-1}\) with respect to \(x\) is:\[ -\left(\frac{d y}{d x}\right)^{-2} \cdot \left(\frac{d^2 y}{d x^2}\right) \]

Step 5: Combine Expressions

Substituting into the expression:\[\frac{d^2 x}{d y^2} = -\left(\frac{d y}{d x}\right)^{-2} \cdot \left(\frac{d^2 y}{d x^2}\right) \cdot \left(\frac{d y}{d x}\right)^{-1} = -\left(\frac{d^2 y}{d x^2}\right) \cdot \left(\frac{d y}{d x}\right)^{-3}\]This matches option (C).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in differential calculus that allows us to differentiate composites of functions. Imagine you want to differentiate a function that is created by plugging one function into another. For example, if you have a function like \( h(x) = f(g(x)) \). Here, \( f \) is applied to \( g(x) \), and to differentiate \( h \) with respect to \( x \), you use the chain rule.

• The formula for the chain rule is: \[ \frac{dh}{dx} = \frac{df}{dg} \times \frac{dg}{dx} \]
• This means you first differentiate \( f \) with respect to \( g \), and then multiply it by the derivative of \( g \) with respect to \( x \).
• This tool is particularly useful when dealing with nested or composite functions.

In the provided problem, we used it to find the derivative of an inverse function. Applying it carefully allows us to navigate the derivatives of complex expressions, especially when functions are intertwined.

Second Order Derivative

The second order derivative gives us information about the curvature or concavity of a function's graph. It is essentially the derivative of the derivative, which is why it's called a second order derivative. For a given function \( y = f(x) \), the process is as follows:

• First, find the first order derivative \( \frac{dy}{dx} \).
• Then, differentiate \( \frac{dy}{dx} \) with respect to \( x \) again to get \( \frac{d^2y}{dx^2} \).
• While the first derivative gives us the slope at any point, the second order derivative informs us whether the slope is increasing or decreasing.

It is widely used in understanding how a function behaves and anticipates inflection points where the graph shifts from concave to convex or vice versa. In the original exercise, we needed to find a second order derivative with respect to a different variable, showcasing how the concept can be extended beyond standard applications.

Inverse Function Derivative

Finding the derivative of an inverse function is a bit trickier than finding for standard functions. Let's say you have a function \( y = f(x) \) which has an inverse \( x = f^{-1}(y) \). You want to find the derivative of this inverse function:

• The key idea is to use the relationship \( \frac{dx}{dy} = \left(\frac{dy}{dx}\right)^{-1} \). This relation holds because dying summed up, the inverse function swaps roles of \( x \) and \( y \).
• This derivative gives how \( x \) changes with a small change in \( y \), essentially expressing the rate of change in terms of the original derivative \( \frac{dy}{dx} \).
• To take it further, when you compute a second order derivative concerning the inverse, you chain the derivatives, applying the chain rule iteratively as seen in the problem solution.

This approach simplifies finding derivatives for inverse functions and ties back to using the chain rule to solve calculus problems involving inverse operations.