Question

Differentiate the following functions: \(y=\frac{x}{1-x^{2}}\)

Short answer

Question: Find the derivative of the function \(y=\frac{x}{1-x^{2}}\). Answer: The derivative of the function \(y=\frac{x}{1-x^{2}}\) is \(y'(x) = \frac{1 + x^2}{(1-x^2)^2}\).

Step-by-step solution

Differentiate u with respect to x

We have \(u = x\). To differentiate \(u\) with respect to \(x\), we have: $$\frac{du}{dx} = 1$$

Differentiate v with respect to x

We have \(v = 1-x^2\). To differentiate \(v\) with respect to \(x\), we have: $$\frac{dv}{dx} = -2x$$

Apply the Quotient Rule

Now that we have the derivatives of \(u\) and \(v\), we can apply the quotient rule to find the derivative of \(y\): $$\frac{dy}{dx} = \frac{\frac{du}{dx}v - u\frac{dv}{dx}}{v^2}$$ Substitute the values of \(u\), \(v\), \(\frac{du}{dx}\), and \(\frac{dv}{dx}\): $$\frac{dy}{dx} = \frac{(1)(1-x^2) - x(-2x)}{(1-x^2)^2}$$

Simplify the expression

Let's simplify the expression obtained in Step 3: $$\frac{dy}{dx} = \frac{1 - x^2 + 2x^2}{(1-x^2)^2}$$ Combine the terms involving \(x^2\) and simplify: $$\frac{dy}{dx} = \frac{1 + x^2}{(1-x^2)^2}$$ So, the derivative of the given function is: $$y'(x) = \frac{1 + x^2}{(1-x^2)^2}$$

Key concepts

Quotient Rule

The Quotient Rule is a fundamental tool in calculus used for finding the derivative of a function that is the ratio of two other functions. Suppose you have a function given in the form \( y = \frac{u}{v} \). Both \( u \) and \( v \) are differentiable functions of \( x \). The Quotient Rule states that the derivative \( \frac{dy}{dx} \) is:

• Take the derivative of \( u \) and multiply it by the function \( v \) (keep it unchanged).
• Subtract the product of \( u \) (unchanged) and the derivative of \( v \).
• Divide the entire expression by \( v^2 \), the square of the function \( v \).

In mathematical notation, this can be written as:\[ \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]This formula allows us to handle functions in fraction form where direct differentiation may be more complex. Always ensure the denominator function \( v \) is not zero, as division by zero is undefined.

Calculus Problems

Calculus problems often involve differentiating complex functions, which can sometimes be intimidating. However, understanding a few key concepts, like the Quotient Rule, can make the process easier:

• Break the problem into smaller parts. Differentiate the numerator and the denominator separately.
• Apply rules like the Quotient Rule systematically.
• After applying the rule, make sure to simplify the expression completely.

Tackling these problems requires practice and a clear understanding of calculus rules. To improve:

• Practice with problems of varying difficulty.
• Check your work by substituting back into the original function to see if it holds.
• Get comfortable with errors — use them as learning tools to understand where mistakes can occur.

Through consistent practice, dealing with calculus problems involving quotients and other complex functions will become more intuitive.

Function Derivative

A derivative is a measure of how a function changes as its input changes. For a given function \( y = f(x) \), the derivative, denoted as \( y'(x) \) or \( \frac{dy}{dx} \), tells us the rate of change or the slope of the function at any point. For example:

• The derivative of a constant function is zero because it does not change.
• For simple functions, like \( y = x^n \), their derivatives follow standard rules, such as the Power Rule: \( \frac{d}{dx}(x^n) = nx^{n-1} \).
• When dealing with more complex functions like quotients or products, specific rules like the Quotient or Product Rule are necessary.

Calculating derivatives correctly is critical since they provide insight into the behavior of functions, including where they increase or decrease or where slopes are steep or flat. Proper understanding of derivatives forms the foundation for solving real-world problems involving rates of change and optimization.