Problem 42 - A Textbook of Differential Calculus

Question

Match the following : \begin{tabular}{l|l} \hline Column I & Column II \\ \hline (A) \(y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\), then \(\frac{d y}{d x}=-\frac{2}{1+x^{2}}\) & (p) for \(x<0\) \\ (B) \(y=\cos ^{-1}\left(\frac{1}{\sqrt{1+x^{2}}}\right)\), then \(\frac{d y}{d x}=\frac{1}{1+x^{2}}\) & (q) for \(x>1\) \\ (C) \(y=\left|e^{|x|}-e\right|\), then \(\frac{d y}{d x}>0\) & (r) for \(x<-1\) \\ (D) \(u=\log |2 x|, v=\left|\tan ^{-1} x\right|\), then \(\frac{d u}{d v}>2\) & (s) for \(-1

Step-by-Step Solution

Answer

(A) with (p), (B) with (q), (C) with (r), and (D) with (s).

Step 1: Understanding the problem

We need to match each item in Column I with the corresponding description in Column II. This involves finding conditions affecting derivatives or expressions in the specified domains.

Step A: Analyzing function (A)

For \( y = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \), differentiate to find \( \frac{dy}{dx} = -\frac{2}{1+x^2} \). This derivative is negative for all \( x \) except \( x = 0 \), but checking carefully, \( \frac{2x}{1+x^2} \) is defined between \(-1\) and \(1\). The derivative matches condition for \( x < 0 \), therefore match (A) with (p).

Step B: Analyzing function (B)

For \( y = \cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) \), differentiate to get \( \frac{dy}{dx} = \frac{1}{1+x^2} \), a positive value. This condition closely matches situations where the variable \( x \) returns a positive result. Specifically, it is suitable for \( x > 1 \), hence match (B) with (q).

Step C: Analyzing function (C)

The function \( y = \left| e^{|x|} - e \right| \) considers the absolute difference from \( e \). As \( e^{|x|} \) increases exponentially, check when \( \frac{dy}{dx} > 0 \). If \( e^{|x|} > e \) increases, this holds true when \( x > 1 \) or \( x < -1 \). Hence, match (C) with (r).

Step D: Analyzing function (D)

For \( u = \log |2x| \) and \( v = |\tan^{-1} x| \), differentiate \( \frac{du}{dv} \) and evaluate conditions. When you differentiate these equations and solve for \( \frac{du}{dv} > 2 \), it is most likely true for \( -1 < x < 0 \) where log changes rapidly compared to arc tangent. Thus, match (D) with (s).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse operations of the trigonometric functions like sine, cosine, and tangent. These functions allow us to determine the angle when given a ratio of sides from a right triangle. For example, the inverse sine function, written as \( \sin^{-1}(x) \), finds an angle \( \theta \) such that \( \sin(\theta)=x \). Similarly, for inverse cosine and tangent, we use \( \cos^{-1}(x) \) and \( \tan^{-1}(x) \) respectively.
Inverse functions are only defined within certain ranges to ensure they are functions; that means for \( \sin \) and \( \cos \), outputs are restricted between \(-1\) and \(1\) for the function to have real values.
These functions form the backbone for trigonometric equations and their derivatives, which often appear in calculus problems. Understanding their properties helps us analyze and differentiate trigonometric expressions in calculus.

Derivative Calculation

Calculating the derivative of a function is a fundamental concept in differential calculus. The derivative provides a measure of how a function changes as its input changes, specifically indicating the rate of change or slope of the function at a specific point. For inverse trigonometric functions, finding derivatives involves using implicit differentiation and recognizing the specific derivative formulas, such as \( \frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}} \) when \( |x|<1 \).
When differentiating more complex expressions like \( \sin^{-1} \left( \frac{2x}{1+x^2} \right) \), the chain rule is often applied. This rule allows us to break a complex problem into simpler parts, tackling each part separately before combining the results. Remember, recognizing derivative patterns of these inverse functions aids in tackling problems involving more complicated calculations.

Function Analysis

Analyzing functions involves understanding their behavior across various intervals, including examining where a function is increasing, decreasing, or remaining constant. When analyzing complex functions, derivatives play a crucial role as they help identify crucial points like maximums, minimums, and inflection points. For instance, in the exercise, the function \( y = \left| e^{|x|} - e \right| \) is analyzed by considering where its derivative is greater than zero, identifying regions in the domain where the function increases.
Through this analysis, intervals are identified where specific statements or conditions are true, allowing us to effectively match functions with their properties or conditions. This type of breakdown is essential in calculus and helps set the foundation for solving optimization problems and understanding overall function behavior throughout different sections of its domain.

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions and are essential in calculus for dealing with rates of growth and decay. The function \( \log_b(x) \) is defined as the power to which the base \( b \) must be raised to produce \( x \). In natural logarithms, the base is \( e \) (approximately 2.718), making \( \log(x) = \ln(x) \).
Understanding the derivative of logarithmic functions is a significant aspect of differential calculus. For example, the derivative of \( \ln(x) \) is given by \( \frac{1}{x} \), which indicates how the rate of change diminishes as \( x \) increases.
In complex expressions, logarithmic differentiation is often used, especially when dealing with products, quotients, or powers. Through this technique, we transform multiplicative complexities into additive simplifications, making differentiation straightforward and manageable.