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How do you find the derivative of the product of two functions that are differentiable at a point?

Short Answer

Expert verified
Answer: The Product Rule is used to find the derivative of the product of two differentiable functions. To find the derivative of the product of two functions f(x) and g(x) using the Product Rule, follow these steps: 1. Identify the given functions f(x) and g(x). 2. Apply the Product Rule formula: h'(x) = f'(x)g(x) + f(x)g'(x), where h(x) = f(x)g(x). 3. Find the derivatives of both f(x) and g(x), denoted as f'(x) and g'(x), respectively. 4. Substitute the derivatives f'(x) and g'(x) into the Product Rule formula to find the derivative of their product. To find the derivative at a specific point x, evaluate f'(x) and g'(x) at that point and substitute into the formula.

Step by step solution

01

Identify the given functions

In this example, we have two functions f(x) and g(x), which are differentiable at a point x. Our goal is to find the derivative of their product, which can be denoted as h(x) = f(x)g(x).
02

Apply the Product Rule

According to the Product Rule, if h(x) = f(x)g(x), then the derivative h'(x) is given by: h'(x) = f'(x)g(x) + f(x)g'(x)
03

Find the derivatives of f(x) and g(x)

To apply the Product Rule formula, we need the derivatives of the given functions f(x) and g(x). This can be achieved by finding f'(x) and g'(x) respectively. In this general case, we will just refer to the derivatives as f'(x) and g'(x).
04

Substitute the derivatives into the Product Rule formula

Now, plug the derivatives of the functions into the formula from step 2: h'(x) = f'(x)g(x) + f(x)g'(x) This gives us the derivative of the product of two functions. To find the derivative of the product at a specific point x, we can evaluate both f'(x) and g'(x) at that point and substitute into the formula.

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Key Concepts

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

Differentiable Functions
Differentiable functions are the cornerstone of calculus, particularly when working with derivatives. A function is considered differentiable at a point if it has a derivative there. What does this mean? Essentially, it implies that the function's graph has a tangent at that point, and therefore, it behaves nicely without any abrupt breaks or sharp turns.

Imagine drawing a curve on a piece of paper—wherever you can lay a ruler tangentially on the curve without any jumps or corners, the function is differentiable at those points. Formally, if \( f(x) \) is differentiable at \( x = a \), then \( f'(a) \) exists, and the graph of \( f(x) \) will have a well-defined tangent line with slope \( f'(a) \) at \( x = a \). This is crucial for applying the product rule as it relies on both functions being differentiable at the point of interest.
Derivative of a Product
When faced with the task of finding the derivative of a product of two functions, we turn to the Product Rule. The Product Rule is a formula that provides a method to differentiate products of functions.

If we denote the functions as \( f(x) \) and \( g(x) \) and their product as \( h(x) = f(x)g(x) \), then, according to the Product Rule, the derivative of the product \( h'(x) \) is given by combining the derivatives of the individual functions:\[ h'(x) = f'(x)g(x) + f(x)g'(x) \]
The significance of the Product Rule cannot be overstated. It allows us to tackle complex expressions where functions are interdependent, like when velocity is a product of time-dependent speed and position functions.
Chain Rule
Another essential tool in our calculus toolbox is the Chain Rule. The Chain Rule is used when we need to differentiate a composite function—a function inside another function. Think Russian nesting dolls, where each doll is a function wrapped around another.

Formally, if you have a composite function \( h(x) = f(g(x)) \), then the derivative of \( h \) with respect to \( x \) is \( h'(x) = f'(g(x)) \cdot g'(x) \). This allows us to peel away the layers, differentiating the outside function first and then multiplying by the derivative of the inner function. It's a cascading chain of derivatives, hence the name.

Understanding when to apply the Chain Rule, as opposed to the Product Rule, is pivotal for students of calculus and underscores the importance of recognizing the type of function at hand before differentiating.
Calculus Education
Calculus is a vast field, and mastering it requires guidance, practice, and a solid grasp of fundamental principles. Calculus education aims to empower students by providing them with the tools and understanding needed to navigate through problems involving change and motion.

A robust calculus education involves interactive learning, practical examples, and a stepwise approach that builds on previous knowledge. Exercises designed for educational impact should include clear explanations of concepts, like differentiable functions, the Product Rule, and the Chain Rule, followed by incremental practice problems that reinforce these concepts.

With the right educational strategies, students can understand and apply concepts such as the Product Rule effectively, making calculus not just a subject to be passed, but a useful skillset to be applied across various real-world scenarios.

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Most popular questions from this chapter

Recall that \(f\) is even if \(f(-x)=f(x),\) for all \(x\) in the domain of \(f,\) and \(f\) is odd if \(f(-x)=-f(x),\) for all \(x\) in the domain of \(f\) a. If \(f\) is a differentiable, even function on its domain, determine whether \(f^{\prime}\) is even, odd, or neither. b. If \(f\) is a differentiable, odd function on its domain, determine whether \(f^{\prime}\) is even, odd, or neither.

Carry out the following steps. a. Verify that the given point lies on the curve. b. Determine an equation of the line tangent to the curve at the given point. $$x^{2}+x y+y^{2}=7 ;(2,1)$$ (Graph cant copy)

Assume \(f\) and \(g\) are differentiable on their domains with \(h(x)=f(g(x)) .\) Suppose the equation of the line tangent to the graph of \(g\) at the point (4,7) is \(y=3 x-5\) and the equation of the line tangent to the graph of \(f\) at (7,9) is \(y=-2 x+23\) a. Calculate \(h(4)\) and \(h^{\prime}(4)\) b. Determine an equation of the line tangent to the graph of \(h\) at \((4, h(4))\)

Tangency question It is easily verified that the graphs of \(y=x^{2}\) and \(y=e^{x}\) have no point of intersection (for \(x>0\) ), while the graphs of \(y=x^{3}\) and \(y=e^{x}\) have two points of intersection. It follows that for some real number \(2 < p < 3,\) the graphs of \(y=x^{p}\) and \(y=e^{x}\) have exactly one point of intersection (for \(x > 0) .\) Using analytical and/or graphical methods, determine \(p\) and the coordinates of the single point of intersection.

Find \(d^{2} y / d x^{2}.\) $$x+y^{2}=1$$

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