Question

Derivative of function \(\mathrm{f}(\mathrm{x})\left[\mathrm{x}^{2} /\left(1+\sin ^{2} \mathrm{x}\right)\right]\) is (a) Even function (b) Odd function (c) Not define (d) Increasing Function

Short answer

The derivative of the function \(f(x) = \frac{x^2}{1+\sin^2 x}\) is \(f'(x) = \frac{2x(1+\sin^2 x) - 2x^2\sin x\cos x}{(1+\sin^2 x)^2}\). After analyzing its properties, we can conclude that the function's derivative is neither an even function, odd function, undefined, nor always increasing. Hence, the correct answer is "None of the above."

Step-by-step solution

Calculate the Derivative of the Function

To find the derivative of the function, \(f(x) = \frac{x^2}{1+\sin^2 x}\), we will make use of the quotient rule, which states that for any two differentiable functions, u(x) and v(x), the derivative of their quotient is given by: \[\frac{d(u/v)}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\] In our case, \(u(x) = x^2\) and \(v(x) =1 + \sin^2(x)\). We first find the derivatives of both functions: \(u'(x) = \frac{d(x^2)}{dx} = 2x\) \(v'(x) = \frac{d(1+\sin^2 x)}{dx} = 2\sin x\cos x\) Now, we apply the quotient rule:

Applying the Quotient Rule

Using the quotient rule, we have: \[f'(x) = \frac{(2x)(1+\sin^2 x) - (x^2)(2\sin x\cos x)}{(1+\sin^2 x)^2}\] Now, let us simplify the expression and analyze the properties of the obtained derivative.

Simplification and Property Analysis

The given function, \(f'(x)\), can be simplified as: \[f'(x) = \frac{2x(1+\sin^2 x) - 2x^2\sin x\cos x}{(1+\sin^2 x)^2}\] Now, let's analyze the properties of this derivative according to the given options: (a) Even function: If \(f'(x) = f'(-x)\), then the function is even. By inserting \(x = -x\), we obtain: \[f'(-x) = \frac{-2x(1+\sin^2 (-x)) + 2x^2\sin (-x)\cos (-x)}{(1+\sin^2 (-x))^2}\] Since \(\sin(-x) = -\sin(x)\) and \(\cos(-x) = \cos(x)\), we can write: \[f'(-x) = \frac{2x(1+\sin^2 x) - 2x^2\sin x\cos x}{(1+\sin^2 x)^2}\] However, \(f'(-x) \neq f'(x)\), so the function is not even. (b) Odd function: If \(f'(-x) = -f'(x)\), then try evaluating it for negative x. \[f'(-x) = -\frac{2x(1+\sin^2 x) - 2x^2\sin x\cos x}{(1+\sin^2 x)^2}\] Now, comparing this with \(f'(x)\), we notice that it is not equal to \(-f'(x)\). Therefore, the function is not odd. (c) Not defined: The function is defined for all values of x and has a continuous domain, so it is not undefined. (d) Increasing Function: If the function's derivative is positive for all values of x, then the function is increasing. We have: \[f'(x) = \frac{2x(1+\sin^2 x) - 2x^2\sin x\cos x}{(1+\sin^2 x)^2}\] The denominator is always positive. However, we cannot determine if the entire expression is positive or negative. Thus, we cannot conclude that the function is always increasing. Based on our analysis, none of the given options (a), (b), (c), or (d) accurately describe the derivative of the function \(f(x) = \frac{x^2}{1 + \sin^2 x}\). The correct answer would be "None of the above."

Key concepts

Derivative Calculations

Finding the derivative of a function is a fundamental concept in calculus. Derivatives represent the rate of change of a function with respect to its variable. In our example, we have the function, \( f(x) = \frac{x^2}{1 + \sin^2 x} \), and we are tasked to compute its derivative.
The derivative is found using specific rules and techniques. For simple functions, basic derivative rules like the power rule, product rule, or chain rule are used. In more complex cases like our function, the quotient rule is helpful.

To start, identify the functions \( u(x) \) and \( v(x) \), which are \( u(x) = x^2 \) and \( v(x) = 1 + \sin^2(x) \) respectively. Next, compute their derivatives:

• \( u'(x) = 2x \)
• \( v'(x) = 2\sin(x)\cos(x) \)

These derivatives are the building blocks to apply further calculus techniques.

Quotient Rule

The quotient rule is used when you need to find the derivative of a division between two functions. It simplifies this process by providing a formula, ensuring you correctly manage the division.

The rule states that if you have two functions \( u(x) \) and \( v(x) \), the derivative of the quotient \( \frac{u(x)}{v(x)} \) is given by:
\[\frac{d(u/v)}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\]
So, apply this formula to our functions:

• \( f'(x) = \frac{(2x)(1+\sin^2 x) - (x^2)(2\sin x\cos x)}{(1+\sin^2 x)^2} \)

The quotient rule helps in stabilizing the calculation complexities involved and is crucial for minimizing errors, especially in complex functions.

Function Properties

Functions have unique properties that can determine their behavior over a range. These properties include being even, odd, or increasing. Knowing these can help you understand their graph patterns and applications.

An **even function** is symmetric around the y-axis, satisfying \( f(x) = f(-x) \). Conversely, an **odd function** has rotational symmetry around the origin, satisfying \( f(-x) = -f(x) \). An **increasing function** is one where the output rises as the input increases.
Analyzing \( f'(x) \) helped us conclude about these properties. However, as explored in our solution:

• \( f'(x) \) is neither even nor odd since the conditions aren't met.
• We cannot definitively tell if it's an increasing function without further inequality analysis.

So, it's essential to test a function's derivative thoroughly to fully understand its overall characteristics.

Even and Odd Functions

Identifying whether a function or its derivative is even or odd helps in predicting the symmetry of its graph. Each form of symmetry gives insights into the function's graphical behavior and can assist in solving problems more efficiently.

To check if a derivative is even, substitute \( x \) with \( -x \) and verify if the expression remains unchanged:

• An even function: \( f(-x) = f(x) \)

For odd functions, substitute \( x \) with \( -x \) and verify if the expression becomes the negative of the original:

• An odd function: \( f(-x) = -f(x) \)

In our example, testing the derivative did not satisfy these conditions. Both checks showed the function behaved neither proportionally with negation nor reflectively, implying the lack of a particular symmetry. Always consider testing these properties to gain deeper insights into function behavior.