Problem 19 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(\phi(x)\) be the inverse of the function \(f(x)\) and \(f^{\prime}(x)=\frac{1}{1+x^{5}}\) then \(\frac{d}{d x} \phi(x)\) is (A) \(\frac{1}{1+[\phi(x)]^{5}}\) (B) \(\frac{1}{1+[f(x)]^{5}}\) (C) \(1+[\phi(x)]^{5}\) (D) \(1+[f(x)]^{5}\)

Step-by-Step Solution

Answer

The correct answer is (C) \(1 + [\phi(x)]^5\).

Step 1: Understand the Given Information

We have a function \( f(x) \) and its derivative \( f'(x) = \frac{1}{1+x^5} \). Additionally, \( \phi(x) \) is the inverse function of \( f(x) \). We need to find the derivative of the inverse function, \( \frac{d}{dx} \phi(x) \).

Step 2: Use Inverse Function Derivative Formula

The derivative of an inverse function \( \phi(x) \) with respect to \( x \) is given by \( \frac{d}{dx} \phi(x) = \frac{1}{f'(\phi(x))} \). We will use this formula to calculate \( \frac{d}{dx} \phi(x) \).

Step 3: Substitute f'(x) in the Formula

Since \( f'(x) = \frac{1}{1+x^5} \), we replace \( f'(\phi(x)) \) with \( \frac{1}{1+[\phi(x)]^5} \) in the derivative formula: \[ \frac{d}{dx} \phi(x) = \frac{1}{\frac{1}{1+[\phi(x)]^5}} \].

Step 4: Simplify the Expression

Simplify the expression from Step 3: \( \frac{1}{\frac{1}{1+[\phi(x)]^5}} = 1 + [\phi(x)]^5 \).

Step 5: Identify the Correct Answer

Comparing \( \frac{d}{dx} \phi(x) = 1 + [\phi(x)]^5 \) with the given options, it matches option (C).

Key Concepts

Derivative

In mathematics, the derivative represents the rate at which a function is changing at any given point. It's akin to the concept of speed or velocity in physics, which tells us how quickly something is moving at a specific instant.

• Notation: Common notations include \( f'(x) \) or \( \frac{df}{dx} \), which denote the derivative of a function \( f \) with respect to \( x \).
• Basic Idea: If you have a graph of a function, the derivative at a particular point is the slope of the tangent line to the function at that point.

To find a derivative, we often use rules such as the power rule, product rule, and chain rule. The derivative can also indicate the nature of the slopes, whether they're increasing or decreasing, helping us understand the general behavior of functions. In this context, the derivative of a function is vital for calculating the rate of change of its inverse.

Inverse Function Theorem

The inverse function theorem is a very useful tool in calculus. It provides a way to find the derivative of an inverse function. When you have a function \( f(x) \) and its inverse \( \phi(x) \), the theorem helps determine the derivative of \( \phi(x) \).

• Theorem Formula: If you have \( f'(x) \), the derivative of its inverse function is \( \frac{1}{f'( ext{inverse})} \). In other words, if \( y = f(x) \), then \( x = f^{-1}(y) \) implies the derivative \( (f^{-1})'(y) = \frac{1}{f'(x)} \).
• Applications: This theorem is particularly useful when direct computation of the inverse is challenging, but you can easily compute the derivative of the original function \( f(x) \).

In our exercise, the theorem is key to solving for \( \frac{d}{dx} \phi(x) \), where \( \phi(x) \) is the inverse of \( f(x) \). Knowing \( f' \) allows us to use the formula to find the desired derivative of the inverse.

Chain Rule

The chain rule is a fundamental technique in calculus used to compute the derivative of a composition of functions. When you have two functions, say \( u(x) \) and \( v(u) \), and you want to find the derivative of \( v(u(x)) \), the chain rule is applied.

• Chain Rule Formula: The chain rule can be expressed as \( \frac{d}{dx}[v(u(x))] = v'(u(x)) \cdot u'(x) \).
• Purpose: It essentially tells us how one function, applied inside another, changes with respect to an external variable.
• Example Use: If you're dealing with nested functions, like \( e^{x^2} \), you first find the derivative of the outer function \( e^u \), then multiply it by the derivative of the inner function \( x^2 \).

In problems involving inverse functions, the chain rule plays a role in simplifying derivatives and understanding relationships between composed functions. By mastering the chain rule alongside the inverse function theorem, you can tackle a wide range of calculus problems effectively.