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Chapter 3: Problem 13

Differentiate the functions. $$y=\frac{x^{2}-1}{x^{2}+1}$$

Short Answer

Expert verified
The derivative is \( y' = \frac{4x}{(x^2 + 1)^2} \).

Step by step solution

01

Identify the Quotient Rule

To differentiate the function \( y = \frac{x^2 - 1}{x^2 + 1} \), notice that it is in the form \(\frac{u}{v}\) where \( u = x^2 - 1 \) and \( v = x^2 + 1 \). We will use the Quotient Rule: \[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \].
02

Differentiate the Numerator and Denominator

First, find the derivatives of the numerator \( u \) and the denominator \( v \). These are: \( u = x^2 - 1 \rightarrow u' = 2x \) \( v = x^2 + 1 \rightarrow v' = 2x \)
03

Apply the Quotient Rule

Now, substitute the derivatives into the Quotient Rule formula: \[ y' = \frac{(2x)(x^2 + 1) - (x^2 - 1)(2x)}{(x^2 + 1)^2} \]
04

Simplify the Numerator

Expand and simplify the expression in the numerator: \[ (2x)(x^2 + 1) - (2x)(x^2 - 1) = 2x^3 + 2x - 2x^3 + 2x \] This simplifies to \[ y' = \frac{2x + 2x}{(x^2 + 1)^2} = \frac{4x}{(x^2 + 1)^2} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation
Differentiation is a crucial concept in calculus. It involves finding the rate at which a function is changing at any given point. When you differentiate a function, you get its derivative, which represents this rate of change. For instance, in the exercise provided, we need to differentiate the function \(\frac{x^2 - 1}{x^2 + 1}\). This process is essential for many applications in science, engineering, and everyday problem-solving.
Derivative
The derivative of a function measures how the function's output changes with respect to changes in its input. In simpler terms, it tells us the slope of the function at any point. For our example, we can use the Quotient Rule to find the derivative of the function \(\frac{x^2 - 1}{x^2 + 1}\). The Quotient Rule states that for functions \(\frac{u}{v}\), where both u and v are differentiable, the derivative is \(\frac{u'v - uv'}{v^2}\). Applying this rule involves carefully differentiating both the numerator and the denominator before combining them using the formula.
Calculus problem-solving
Solving calculus problems often involves breaking down the problem into smaller, manageable steps. This systematic approach makes complex problems easier to tackle. In our example, we first identify the forms of u and v, find their derivatives, and then apply the Quotient Rule. Finally, we simplify the resulting expression to obtain the derivative. This structured problem-solving method is something you can apply to various calculus problems, helping you to understand the underlying concepts deeply and efficiently.

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Most popular questions from this chapter

When a company produces and sells \(x\) thousand units per week, its total weekly profit is \(P\) thousand dollars, where $$ P=\frac{200 x}{100+x^{2}} . $$ The production level at \(t\) weeks from the present is \(x=4+2 t\) (a) Find the marginal profit, \(\frac{d P}{d x}\). (b) Find the time rate of change of profit, \(\frac{d P}{d t}\). (c) How fast (with respect to time) are profits changing when \(t=8 ?\)

Suppose that \(x\) and \(y\) are related by the given equation and use implicit differentiation to determine \(\frac{d y}{d x}\). $$x^{2} y^{3}=6$$

Suppose that \(x\) and \(y\) are both differentiable functions of \(t\) and are related by the given equation. Use implicit differentiation with respect to \(t\) to determine \(\frac{d y}{d t}\) in terms of \(x, y\), and \(\frac{d x}{d t}\). $$y^{2}=8+x y$$

Compute \(\frac{d}{d x} f(g(x))\), where \(f(x)\) and \(g(x)\) are the following: $$f(x)=\sqrt{x}, g(x)=x^{2}+1$$

Suppose that the price \(p\) (in dollars) and the demand \(x\) (in thousands of units) of a commodity satisfy the demand equation $$ 6 p+x+x p=94 $$ How fast is the demand changing at a time when \(x=4\), \(p=9\), and the price is rising at the rate of $$\$ 2$$ per week?
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