Question

True or False? In Exercises 137-139, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(y\) is a differentiable function of \(u, u\) is a differentiable function of \(v,\) and \(v\) is a differentiable function of \(x,\) then \(\frac{d y}{d x}=\frac{d y}{d u} \frac{d u}{d v} \frac{d v}{d x}\)

Short answer

The statement is true. It is a correct application of the chain rule for three composite functions.

Step-by-step solution

Recall or Look Up the Chain Rule

The chain rule states that if we have a composite function, the derivative of the outer function is taken first, then multiplied by the derivative of the inner function. For three composite functions \(u(v(x))\) and \(y(u(v(x)))\), the chain rule states that $\frac{d y}{d x}=\frac{d y}{d u} \frac{d u}{d v} \frac{d v}{d x}$. Knowing this is critical to proceeding with the exercise.

Compare the Given Statement with the Chain Rule

The formula given in the exercise is indeed the expression of the chain rule for the composition of three functions.

Statement Verification

Since the given statement matches the chain rule for the composition of three functions, the statement is true.

Key concepts

Differentiable Function

To understand the depth of calculus, grasping the concept of a differentiable function is vital. It refers to a function that has a derivative at each point in its domain. Simply put, if you can draw a tangent line at any point along the curve of the function, then it's differentiable at that point. Here's why this is important: differentiation is the process of finding the rate at which one quantity changes with respect to another, and if a function is not differentiable, we can't proceed with finding its derivative.
In practical terms, if a function has any sharp corners, discontinuities, or vertical tangents, it's not differentiable at those points. For example, the absolute value function, written as \( f(x) = |x| \), has a sharp corner at the point \( x=0 \), hence it's not differentiable at that point.
Differentiability is a cornerstone in calculus because it ensures the continuity of functions, and it's assumed in most calculus problems. It's also the prerequisite for using the chain rule, which is a method used to find derivatives of composite functions.

Composite Function

Diving further into calculus, we encounter composite functions, which are essentially functions made up by combining two or more functions. Here's a simple way to think about it: if you have two functions, let's call them \( f \) and \( g \), a composite function \( h \) is created when you apply \( g \) first, then apply \( f \) to the result. In notation, we write \( h(x) = f(g(x)) \).
Why does this matter? Composite functions are ubiquitous in mathematics and the real world. For instance, if you're calculating the area of a painted surface where the rate of painting depends on time, you're potentially dealing with a composite function: the area depends on the rate, which depends on time.
A key aspect of working with composite functions is understanding how to find their derivatives, which is where the chain rule comes into play. Calculus problems often require unraveling these layers of functions to find the rate of change at a deep level.

Derivative

The concept of derivative is arguably the most central in calculus. In the simplest terms, the derivative of a function represents the rate of change of a quantity with respect to another. If you're plotting the position of a car over time, the derivative gives you the speed—it tells you how fast the car's position is changing over time.
Mathematically, the derivative of a function \( f \) at a point \( x \) is the limit of the average rate of change of the function between two points as the distance between the points approaches zero. This can be symbolized as:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
In practice, finding derivatives can become complex, particularly for composite functions. This necessitates a strong understanding of both the concept and the techniques for finding derivatives—such as the chain rule, which is a powerful tool in the calculus toolkit for handling more sophisticated scenarios.