Question

Product Rule for three functions Assume that $f,g,$ and $h$ are differentiable at $x$ a. Use the Product Rule (twice) to find a formula for $\frac{d}{dx}(f(x)g(x)h(x))$ b. Use the formula in (a) to find $\frac{d}{dx}(e^{2x}(x-1)(x+3))$

Short answer

Question: Find the derivative of the function $e^{2x}(x-1)(x+3)$ using the general formula for the derivative of the product of three functions. Answer: The derivative of the function $e^{2x}(x-1)(x+3)$ is $(2e^{2x}(x-1)+e^{2x})(x+3)+e^{2x}(x-1)$.

Step-by-step solution

Finding the general formula for the derivative of the product of three functions

Recall the Product Rule: $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$. Our goal is to find the derivative of $f(x)g(x)h(x)$. Now consider the product of $f(x)$ and $g(x)$ as $p(x)=f(x)g(x)$. Applying the Product Rule, we get: $$p^{\prime }(x)=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$$ Now, let's consider the derivative of the product $p(x)h(x)$. Apply the Product Rule for this: $$\frac{d}{dx}(f(x)g(x)h(x))=(p^{\prime }h(x)+p(x)h^{\prime }(x))$$ Substitute the expression for $p^{\prime }(x)$ found earlier to get the general formula: $$\frac{d}{dx}(f(x)g(x)h(x))=(f^{\prime }(x)g(x)+f(x)g^{\prime }(x))h(x)+f(x)g(x)h^{\prime }(x)$$ This is the desired general formula.

Applying the general formula to find the derivative of the given function

Now, let's use the found formula to find the derivative of the given function: $e^{2x}(x-1)(x+3)$. First, let's identify the functions: $f(x)=e^{2x}$ $g(x)=x-1$ $h(x)=x+3$ Now, we need to find the derivatives: $f^{\prime }(x)=\frac{d}{dx}(e^{2x})=2e^{2x}$ $g^{\prime }(x)=\frac{d}{dx}(x-1)=1$ $h^{\prime }(x)=\frac{d}{dx}(x+3)=1$ Now, follow the general formula to find the derivative: $$\frac{d}{dx}(f(x)g(x)h(x))=(f^{\prime }(x)g(x)+f(x)g^{\prime }(x))h(x)+f(x)g(x)h^{\prime }(x)$$ Plug in the values for $f(x),f^{\prime }(x),g(x),g^{\prime }(x),h(x),$ and $h^{\prime }(x)$. Calculate the result: $$\frac{d}{dx}(e^{2x}(x-1)(x+3))=(2e^{2x}(x-1)+e^{2x}(1))(x+3)+e^{2x}(x-1)(1)$$ So, the derivative we were seeking is: $$\frac{d}{dx}(e^{2x}(x-1)(x+3))=(2e^{2x}(x-1)+e^{2x})(x+3)+e^{2x}(x-1)$$

Key concepts

Derivative

Understanding derivatives is a fundamental concept in calculus. When you think of a derivative, imagine it as the rate at which a function changes. To put it simply, a derivative tells you how fast or slow a function is increasing or decreasing at any given point.
In mathematical terms, the derivative of a function $f(x)$ with respect to $x$ is represented as $f^{\prime }(x)$ or $\frac{df}{dx}$. This represents the slope of the tangent line to the curve of the function at any point.
- It's like finding the instantaneous speed if the function was a graph of position over time.- The process of finding a derivative is known as differentiation.
Differentiation can involve several rules like the Power Rule, Product Rule, and Chain Rule, which help in finding derivatives of more complex functions. In exercises, these rules often come into play, especially when dealing with products or compositions of functions.

Differentiable Functions

A differentiable function is one where the derivative exists at every point in its domain. This means that the function is smooth without any sharp corners or discontinuities.
Think of a differentiable function as having a predictable pattern without any breaks. This characteristic is crucial because differentiated functions allow us to apply derivative rules and make predictions about the function's behavior.
- If a graph has a sharp turn or a cusp, the derivative does not exist at that point, thus making it non-differentiable at that point. - Differentiability implies continuity – but not all continuous functions are differentiable.
In the context of exercises, when we assume functions are differentiable, we're saying they are smooth enough for us to apply calculus rules like the Product Rule accurately without encountering undefined points.

Calculus

Calculus is one of the most significant branches of mathematics, focusing on change and motion. It is divided into two main parts: differential calculus and integral calculus.
- **Differential Calculus**: This deals with the concept of the derivative. It helps us understand rates of change and slopes of curves. - **Integral Calculus**: This complements the differential side by focusing on accumulation of quantities, such as areas under curves.
The Product Rule, which is used in the exercise above, is part of differential calculus. It allows us to find the derivative of the product of two or more functions. This is especially useful when dealing with real-world problems involving multiple varying quantities.
- Calculus provides tools to model and solve problems in physics, engineering, economics, and more. - Understanding its concepts opens up a world of analytical power to solve practical problems and perform sophisticated analysis.