Question

Assume \(f, g,\) and \(h\) are differentiable at \(x.\) a. Use the Product Rule (twice) to find a formula for \(d / d x(f(x) g(x) h(x)).\) b. Use the formula in (a) to find \(d / d x\left(e^{x}(x-1)(x+3)\right).\)

Short answer

Question: Find the derivative of the function e^x(x-1)(x+3). Answer: The derivative of the function e^x(x-1)(x+3) is: $$ \frac{dy(x)}{dx} = e^{x}(x-1) + e^{x}(x+3) + e^{x}(x-1)(x+3) $$

Step-by-step solution

Apply the Product Rule to find the derivative of f(x)g(x)h(x)

To find the derivative of the product of three functions, we can treat two of the functions as a single function and apply the product rule. Let's consider the product of f(x) and g(x) as a single function, say u(x) = f(x)g(x). Then, our original function becomes the product of two functions: u(x)h(x). Now, we can apply the product rule. The product rule states that if two functions u(x) and v(x) are differentiable, then the derivative of their product is given by: $$ \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} $$ So, the derivative of u(x)h(x) with respect to x is: $$ \frac{d(uh)}{dx} = u(x) \frac{dh(x)}{dx} + h(x) \frac{du(x)}{dx} $$ Since u(x) = f(x)g(x), we now have: $$ \frac{d(fg h)}{dx} = f(x)g(x) \frac{dh(x)}{dx} + h(x) \frac{d(f(x)g(x))}{dx} $$ Now, we need to find the derivative of f(x)g(x) using the product rule.

Apply the Product Rule to find the derivative of f(x)g(x)

Applying the product rule on f(x) and g(x) gives: $$ \frac{d(fg)}{dx} = f(x) \frac{dg(x)}{dx} + g(x) \frac{df(x)}{dx} $$ Now, we can replace this derivative back into our previous equation: $$ \frac{d(fgh)}{dx} = f(x)g(x) \frac{dh(x)}{dx} + h(x) \left(f(x) \frac{dg(x)}{dx} + g(x) \frac{df(x)}{dx}\right) $$ This is the formula for the derivative of a product of three functions.

Apply the formula to find the derivative of e^x(x-1)(x+3)

Let's now use the formula we obtained in part a to find the derivative of the given function: $$ y(x) = e^{x}(x-1)(x+3) $$ Here, we have f(x) = e^x, g(x) = (x - 1), and h(x) = (x + 3). Now, we can find their derivatives: $$ \frac{df(x)}{dx} = e^{x}, \frac{dg(x)}{dx} = 1, \frac{dh(x)}{dx} = 1 $$ Now, we can plug these functions and their derivatives into our formula: $$ \frac{dy(x)}{dx} = e^{x}(x-1)\left(\frac{d(x+3)}{dx}\right) + (x+3)\left(e^{x}\left(\frac{d(x-1)}{dx}\right) + (x-1)\left(\frac{d e^x}{dx}\right)\right) $$ Which simplifies to: $$ \frac{dy(x)}{dx} = e^{x}(x-1)(1) + (x+3)\left(e^{x}(1) + (x-1)(e^{x})\right) $$ Finally, we have the derivative of the given function: $$ \frac{dy(x)}{dx} = e^{x}(x-1) + e^{x}(x+3) + e^{x}(x-1)(x+3) $$

Key concepts

Product Rule

The Product Rule is a fundamental theorem in calculus that allows us to compute the derivative of a product of two or more differentiable functions. According to this rule, if you have functions u(x) and v(x) that are differentiable, the derivative of their product u(x)v(x) is given by:

\[\begin{equation}\frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}\end{equation}\]
This means that you take the derivative of the first function, multiply it by the second function as it is, then add the product of the first function as it is with the derivative of the second function. To solve more complex products involving three or more functions, we can apply the Product Rule iteratively. That is, we first apply the Product Rule to two of the functions and treat the result as one function in a subsequent Product Rule application.

Differentiable Functions

Differentiability is a cornerstone concept in calculus. A function is considered differentiable at a point if it has a derivative at that point. Graphically, this implies there is a tangent line at that point on the curve. Mathematically, for a function to be differentiable at point x, the limit:\[\begin{equation}\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\end{equation}\]
must exist. Differentiable functions are smooth and continuous, meaning there are no sharp corners or breaks in their graphs. When dealing with product and chain rules, it is essential that the functions involved are differentiable, as these rules rely on the functions' smooth behavior to compute derivatives accurately.

Calculus

Calculus is a branch of mathematics that deals with rates of change (differential calculus) and the accumulation of quantities (integral calculus). It provides a framework for modeling and solving problems involving dynamic systems, growth and decay, motion, and many other real-world phenomena. Calculus allows us to find the slopes of curves, compute areas under curves, and much more. Whether using the Product Rule to find the derivative of a function or applying the Chain Rule to compose derivatives, calculus equips students with the tools to tackle complex mathematical challenges.

Chain Rule

The Chain Rule is another essential rule in differential calculus, primarily used when dealing with the derivative of a composite function. If we have a function h(x) = f(g(x)), then the derivative of h with respect to x is the product of the derivative of f with respect to g(x) and the derivative of g with respect to x:\[\begin{equation}\frac{dh}{dx} = \frac{df}{dg} \times \frac{dg}{dx}\end{equation}\]
When using the Product Rule for the derivative of a product of multiple functions, we may need to use the Chain Rule to differentiate composite components within the product. Understanding both the Product and Chain Rules is crucial for students to navigate more complex differentiation tasks.