Problem 56 - Thomas Calculus

Question

Derivative of multiples Does knowing that a function \(g(t)\) is differentiable at \(t=7\) tell you anything about the differentiability of the function 3\(g\) at \(t=7 ?\) Give reasons for your answer.

Step-by-Step Solution

Answer

Yes, if \( g(t) \) is differentiable at \( t=7 \), then \( 3g(t) \) is also differentiable there by the constant multiple rule.

Step 1: Understand the Problem

We need to determine whether the differentiability of a function \( g(t) \) at a specific point \( t = 7 \) has any implications for the differentiability of the function \( 3g(t) \) at the same point.

Step 2: Review the Properties of Derivatives

Recall that if a function \( g(t) \) is differentiable, it means \( g'(t) \) exists at the point of interest. Differentiability implies continuity at that point as well.

Step 3: Apply the Constant Multiple Rule for Derivatives

The constant multiple rule of differentiation states that if \( g(t) \) is differentiable at \( t \), then for any constant \( c \), \( c \cdot g(t) \) is also differentiable at \( t \) with the derivative \( (c \cdot g)'(t) = c \cdot g'(t) \).

Step 4: Analyze the Given Functions

In this case, the function is \( 3g(t) \), where the constant \( c = 3 \). Applying the constant multiple rule, since \( g(t) \) is differentiable at \( t = 7 \), \( 3g(t) \) must be differentiable at \( t = 7 \) as well.

Key Concepts

Constant Multiple Rule

To start, let's understand the Constant Multiple Rule in derivatives. When you're working with differentiable functions, this rule is incredibly useful. It states that if you have a function \( g(t) \) which is differentiable at a point \( t \), then multiplying this function by a constant \( c \) also results in a differentiable function at that same point.
The beauty of this rule lies in its simplicity. It allows us to scale functions without losing differentiability. Here's how it works: The derivative of the product \( c \cdot g(t) \) is simply the constant \( c \) times the derivative of \( g(t) \). Mathematically, this is expressed as:

• \((c \cdot g)'(t) = c \cdot g'(t)\)

This principle is essential in calculus because it means that as long as the original function \( g(t) \) is differentiable, scaling it by any constant \( c \) keeps it differentiable. Thus, in our problem, the differentiability of \( g(t) \) at \( t = 7 \) guarantees the differentiability of \( 3g(t) \) at \( t = 7 \) as well. This constant multiplier does not alter the fact that we can derive the function normally.

Differentiability

Differentiability is a core concept in calculus that refers to the existence of a derivative for a function at a particular point. If a function \( g(t) \) is differentiable at a given point, it means that it has a defined slope, or rate of change, at that point. In simpler terms, the function is smooth and has no sharp corners or discontinuities there.
A function's differentiability is not something we assume; it needs to be verified through the mathematical definition of a derivative. For a function to be differentiable at a point \( t \), the limit that defines the derivative must exist. This is expressed as:

• \( g'(t) = \lim_{{h \to 0}} \frac{g(t+h) - g(t)}{h} \)

If this limit exists, we say that \( g(t) \) is differentiable at \( t \). Being differentiable implies continuity but requires more; the rate of change must not just exist but be finite and well-defined. In our problem, knowing \( g(t) \) is differentiable at \( t = 7 \) directly informs us that operations such as multiplying by a constant keep it differentiable, assuring us of smoothness and the ability to compute derivatives consistently.

Continuity

Continuity is another fundamental concept in calculus, closely related to differentiability. In essence, a function \( g(t) \) is continuous at a point \( t \) if there are no abrupt changes in its value as \( t \) approaches that point. This means you can draw the function at that point without lifting your pencil from the paper.
Mathematically, a function is continuous at \( t \) if the following holds:

• \( \lim_{{x \to t}} g(x) = g(t) \)

One key takeaway here is that if a function is differentiable at \( t \), it must also be continuous at \( t \). However, the converse isn't true; a continuous function isn't necessarily differentiable. It can have sharp turns or vertical tangents.
In the context of our specific problem, \( g(t) \) being differentiable at \( t = 7 \) means it is definitely continuous at that point. Therefore, the function \( 3g(t) \) is also continuous at \( t = 7 \) due to the multiplication by a constant not affecting continuity. Differentiability requires this step, ensuring that we can handle more advanced calculus operations securely.