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Quotient Rule for the second derivative Assuming the first and second derivatives of \(f\) and \(g\) exist at \(x,\) find a formula for \(\frac{d^{2}}{d x^{2}}\left(\frac{f(x)}{g(x)}\right)\)

Short Answer

Expert verified
Answer: The formula for the second derivative of the quotient \(\frac{f(x)}{g(x)}\) with respect to \(x\) is \(\frac{d^{2}}{dx^2}\left(\frac{f(x)}{g(x)}\right) = \frac{f''(x)g^2(x) - 2f'(x)g(x)g'(x) + f(x)g^2(x)g''(x) - f(x)g^4(x)}{g^4(x)}\).

Step by step solution

01

Apply the Quotient Rule for the First Derivative

The Quotient Rule states that the derivative of a quotient \(\frac{f(x)}{g(x)}\) is given by: \(\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g^2(x)}\) Apply the Quotient Rule to our given function: \(\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g^2(x)}\)
02

Apply the Quotient Rule again for the Second Derivative

Now, we need to find the derivative of the expression we obtained in step 1: \(\frac{d^{2}}{dx^2} \left(\frac{f(x)}{g(x)}\right) = \frac{d}{dx} \left(\frac{f'(x)g(x) - f(x)g'(x)}{g^2(x)}\right)\) Applying the Quotient Rule again, we get: \(\frac{d^{2}}{dx^2} \left(\frac{f(x)}{g(x)}\right) = \frac{\Big(\frac{d}{dx}(f'(x)g(x) - f(x)g'(x))\Big)g^2(x) - \Big(\frac{d}{dx}g^2(x)\Big)(f'(x)g(x) - f(x)g'(x))}{\Big(g^2(x)\Big)^2}\)
03

Simplify the expression

To simplify, we'll use the product rule to find the derivatives in the numerator: \(\frac{d}{dx}(f'(x)g(x)) = f''(x)g(x) + f'(x)g'(x)\) \(\frac{d}{dx}(f(x)g'(x)) = f'(x)g'(x) + f(x)g''(x)\) \(\frac{d}{dx}g^2(x) = 2g(x)g'(x)\) Plug these into the expression we obtained in step 2: \(\frac{d^{2}}{dx^2} \left(\frac{f(x)}{g(x)}\right) = \frac{\big(f''(x)g(x) + f'(x)g'(x) - f'(x)g'(x) - f(x)g''(x)\big)g^2(x) - 2g(x)g'(x)(f'(x)g(x) - f(x)g'(x))}{g^4(x)}\) Now, simplify the expression: \(\frac{d^{2}}{dx^2} \left(\frac{f(x)}{g(x)}\right) = \frac{f''(x)g^2(x) - 2f'(x)g(x)g'(x) + f(x)g^2(x)g''(x) - f(x)g^4(x)}{g^4(x)}\) Thus, the second derivative of \(\frac{f(x)}{g(x)}\) with respect to \(x\) is: \(\frac{d^{2}}{dx^2}\left(\frac{f(x)}{g(x)}\right) = \frac{f''(x)g^2(x) - 2f'(x)g(x)g'(x) + f(x)g^2(x)g''(x) - f(x)g^4(x)}{g^4(x)}\)

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Key Concepts

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

Derivative
A derivative provides a way to understand how a function changes as its input changes. When you hear the term "derivative," think of it as the rate of change or the slope of a function at any given point. The concept is a cornerstone of calculus and is used to find the slope of a tangent line to a curve at a point.
The derivative of a function, denoted as \(f'(x)\), tells us how the value of the function \(f(x)\) changes with a slight change in \(x\). In practical terms, it can tell you the speed of a moving object if the position of the object is given as a function of time. The derivative thus captures the idea of instantaneously changing rates that are essential in more complex calculations and problems like optimization and motion analysis.
In the context of the exercise given, the derivative of the quotient \(\frac{f(x)}{g(x)}\) was found using the Quotient Rule. The process involved finding the derivative of the top and bottom functions separately and combining them per the rule's formula. This rule is crucial when dealing with problems where functions are divided by each other.
Second Derivative
The second derivative, denoted as \( \frac{d^2}{dx^2} f(x) \), takes the idea of a derivative one step further. While the first derivative \( f'(x) \) gives the rate of change of \( f(x) \), the second derivative gives the rate of change of the rate of change—that is, it describes how the rate of change itself varies. In physical terms, if the first derivative represents velocity, the second derivative represents acceleration.
The second derivative can inform us whether the rate of change of a function is increasing or decreasing, and it can also provide insight into concavity, which tells us if a curve is bending upwards or downwards. This is crucial for analyzing and predicting the behavior of figures in various science and engineering fields.
In solving the original exercise, you find the second derivative of the quotient function by applying the Quotient Rule again to the first derivative result. It’s essential to simplify the resultant expression for practical use, ensuring it can easily describe the behavior of dynamic systems.
Product Rule
The Product Rule is a vital tool in calculus used when taking derivatives of products of two functions. Unlike simpler derivatives, where basic rules apply, the derivative of the product of two functions involves more work. The Product Rule states that if you have functions \(u(x)\) and \(v(x)\), the derivative of their product is given by \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \).
It ensures you capture the essence of how each function contributes to the product's rate of change. This rule becomes especially useful when dealing with physics problems where quantities are interdependent or tied together, like work done over time, which might involve both force and distance functions.
In the original exercise, the Product Rule was used to simplify the expression for the second derivative by finding the derivative of products inside the quotient expression. Knowing when and how to apply the Product Rule can significantly simplify complex derivative problems.

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Most popular questions from this chapter

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