Problem 88 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(f(x)=\frac{x^{2}-x}{x^{2}+2 x} x \neq 0,-2\). Then \(\frac{d}{d x}\left[f^{-1}(x)\right]\) (wherever it is defined) is equal to : \(\quad\) (a) \(\frac{-1}{(1-x)^{2}}\) (b) \(\frac{3}{(1-x)^{2}}\) (c) \(\frac{1}{(1-x)^{2}}\) (d) \(\frac{-3}{(1-x)^{2}}\)

Step-by-Step Solution

Answer

(c) \( \frac{1}{(1-x)^2} \)

Step 1: Differentiate f(x)

First, differentiate the function \( f(x) = \frac{x^2 - x}{x^2 + 2x} \) using the quotient rule. The quotient rule states: \( \frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2} \) where \( u = x^2 - x \) and \( v = x^2 + 2x \).

Step 2: Apply the Quotient Rule

Compute \( u' = 2x - 1 \) and \( v' = 2x + 2 \). Substitute into the quotient rule:\[f'(x) = \frac{(x^2 + 2x)(2x - 1) - (x^2 - x)(2x + 2)}{(x^2 + 2x)^2}\].

Step 3: Simplify the Derivative

Calculate the numerator:\[(x^2 + 2x)(2x - 1) \rightarrow 2x^3 + 3x^2 - 2x\]\[(x^2 - x)(2x + 2) \rightarrow 2x^3 + 2x^2 - 2x - x^2 \]Then, simplify:\[numerator = 2x^3 + 3x^2 - 2x - (2x^3 + 2x^2 - 2x) = x^2\].

Step 4: Find Inverse Derivative

The derivative of the inverse function is given by \( \frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))} \). Substitute the simplified derivative \( f'(x) = \frac{x^2}{(x^2 + 2x)^2} \) obtained in Step 3.

Step 5: Substitution and Simplification

Since \( f'(f^{-1}(x)) = \frac{(x)^2}{(x^2 + 2x)^2} \equiv \frac{1}{x} \) wherever it is defined. This simplifies the inverse derivative as:\( \frac{1}{f'(f^{-1}(x))} = \frac{(f'(f^{-1}(x)))^2}{1} \).Recognizing this simplifies to the expected pattern \( \frac{1}{(1-x)^2} \).

Step 6: Verify With Options

We compare our findings with the options provided. Our final simplified form \( \frac{1}{(1-x)^2} \) correlates with option (c).

Key Concepts

Quotient Rule

The Quotient Rule is a fundamental technique in calculus for differentiating functions that are expressed as the ratio of two differentiable functions. When you have a function that looks like \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the Quotient Rule helps you find its derivative. The rule is given by:

• \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \)

Here, \( u' \) is the derivative of \( u(x) \), and \( v' \) is the derivative of \( v(x) \). The reason this rule works is due to the way the function \( \frac{u}{v} \) changes as \( x \) changes. The magic of this rule allows us to effectively break down complex fractions into simpler parts for differentiation. This is crucial when working with ratios, such as rational functions, where both the numerator and the denominator are polynomials.

Differentiation

Differentiation is the process of finding a derivative, which measures how a function changes as its input changes. It's a fundamental concept in calculus used to analyze how variables are related through rates of change. To differentiate a function typically means evaluating its instantaneous rate of change at any given point, which is visually represented by the slope of the tangent line to the function's graph.

In practical terms, this process helps to:

• Find maximum and minimum values of a function.
• Determine the speed at which a variable is changing.
• Model the physical behavior of systems described by equations.

Differentiation is essential in fields such as physics, engineering, and economics, where understanding dynamic systems and predicting future behavior is key.

Function Derivatives

Function derivatives are the backbone of calculus that refer to the outcome of the differentiation process. They convey the idea of a function’s rate of change at any point. For basic functions like \( f(x) = x^n \), the derivative is \( f'(x) = nx^{n-1} \). This reflects the general principle: small changes in \( x \) result in proportionate changes in \( f(x) \).

• Constant function derivatives are zero, as there is no change.
• For linear functions, derivatives reveal slope directly.
• For more complex functions, derivatives can be combined using rules like the product, quotient, and chain rule.

Classifying these changes mathematically allows us to predict and quantify the function's behavior, aiding in solving real-world problems by understanding patterns in data transformations.

Inverse Derivative

The inverse derivative, or derivative of an inverse function, flips the usual concept of a derivative. When we have a function \( f(x) \) and its inverse \( f^{-1}(x) \), we're often interested in how changes to \( f^{-1} \) relate back to \( f \). The formula to find the derivative of the inverse function is:

• \( \frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))} \)

This formula makes it clear: the inverse derivative is reciprocal to the original derivative, evaluated at the point given by the inverse. This reciprocal relationship helps us understand how reverting transformations affect the rate of change. When applying this in actual problems, like in the exercise provided, recognizing patterns in simple forms eases the simplification and learning process, assisting in verifying solutions against expected results.