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Use the calculation thought experiment to say whether the expression is written as a sum, difference, scalar multiple, product, or quotient. Then use the appropriate rules to find its derivative. HINT [See Quick Examples on page 814 and Example 7.] \(y=\frac{(x+2) x}{x+1}\) (Do not simplify the answer.)

Short Answer

Expert verified
The given expression is a quotient. Its derivative, using the Quotient Rule, is \(\frac{dy}{dx} = \frac{(2x + 2)(x + 1) - (x + 2)x}{(x + 1)^2}\) without simplification.

Step by step solution

01

1. Identify the numerator and denominator

In the expression \(y = \frac{(x + 2)x}{x + 1}\), we can see that the numerator \(u(x) = (x + 2)x\) and the denominator \(v(x) = x + 1\).
02

2. Calculate the derivative of the numerator

We will find the derivative of the numerator function \(u(x) = (x + 2)x\). Using the Product Rule, we have: \(u'(x) = \frac{d}{dx}[(x + 2)x] = (x + 2)\frac{d}{dx}(x) + x\frac{d}{dx}(x + 2)\) Here, we have \(\frac{d}{dx}(x) = 1\) and \(\frac{d}{dx}(x + 2) = 1\). Substituting these values, we get: \(u'(x) = (x + 2)(1) + x(1) = x + 2 + x\) So, \(u'(x) = 2x + 2\).
03

3. Calculate the derivative of the denominator

Now, we need to find the derivative of the denominator function \(v(x) = x + 1\): \(v'(x) = \frac{d}{dx}(x + 1)\) Since \(\frac{d}{dx}(x) =2 1\) and \(\frac{d}{dx}(1) = 0\), we have: \(v'(x) = 1 + 0 = 1\)
04

4. Apply the Quotient Rule

Now we will apply the Quotient Rule to find the derivative of the given expression. The Quotient Rule states that if \(y = \frac{u(x)}{v(x)}\) where \(u(x)\) and \(v(x)\) are differentiable functions, then: \(\frac{dy}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}\) So, using the derivatives of the numerator and the denominator we calculated earlier, we get: \(\frac{dy}{dx} = \frac{(2x + 2)(x + 1) - (x + 2)x}{(x + 1)^2}\) This is the derivative of the given expression without simplifying.

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Key Concepts

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

Derivative of a Function
Understanding the derivative of a function is fundamental in calculus, as it measures how a function changes as its input changes. It's like capturing a snapshot of the moment when you're driving a car and looking at the speedometer. That instant speed is what a derivative is for functions.

The derivative is represented as \(\frac{df}{dx}\) or \(f'(x)\), where \(f\) is a function and \(x\) is the variable. In simpler terms, if you have a function \(y = f(x)\), its derivative \(f'(x)\) tells you the slope of the tangent line to the graph of the function at any point \(x\). This slope reflects how fast \(y\) is changing with respect to \(x\).

For basic functions, such as linear functions (like \(y = mx+b\)) where the graph is a straight line, the derivative is constant. But for more complex functions – including polynomials, trigonometric, exponential, and logarithmic functions – the derivative can vary at different points, indicating how the function is 'curved' at those points.

Calculating Simple Derivatives

For the power of \(x\), such as \(x^n\), the derivative is \(nx^{n-1}\), following the power rule. And for the sum or difference of functions, the derivative is simply the sum or difference of their individual derivatives.
Product Rule
When it comes to multiplying two functions, calculus doesn't want you to be left in the dark. Here comes the product rule to the rescue! The product rule is a formula used to find the derivative of a product of two functions. It's like a dance between two parts of a function that are being multiplied together.

If you have two differentiable functions \(u(x)\) and \(v(x)\), and you need to find the derivative of their product \(u(x)v(x)\), simply follow this choreography: \(u'(x)v(x) + u(x)v'(x)\), where \(u'(x)\) and \(v'(x)\) are the derivatives of \(u(x)\) and \(v(x)\), respectively.

In our exercise, we applied the product rule to find the derivative of \((x + 2)x\). We differentiated each function — \(x+2\) and \(x\) — and added the product of the first function and the derivative of the second to the product of the second function and the derivative of the first. It’s a tango of differentiation and multiplication being performed accordingly to the rhythm of the rule.
Quotient Rule
The quotient rule is another essential move in the calculus dance, particularly when you're dividing two functions. It tackles the derivative of a division of two functions, similar to how the product rule handles multiplication.

Thinking in terms of our exercise, when the function you’re dealing with looks like a fraction, the quotient rule is your go-to. The rule states: if you have a function \(y = \frac{u(x)}{v(x)}\), then the derivative of \(y\) with respect to \(x\) is \(\frac{dy}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}\). Here, you subtract the product of the derivative of the numerator and the denominator from the product of the numerator and the derivative of the denominator, all over the square of the denominator.

Why all this fuss? Well, imagine trying to see how the ratio of two changing quantities changes. The quotient rule gives you just that, reflection on instant changes of rates. Always remember that when using the quotient rule, the function on the bottom of the fraction should never equal zero, as dividing by zero is undefined.

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