Question

Taking for granted that \(e^{t}\) is its own derivative, use the chain rule to find \(d y\) jd for the following: (a) \(y=e^{5 t}\) (b) \(y=4 e^{3 t}\) (c) \(y=6 e^{-2 t}\)

Short answer

(a) \(5e^{5t}\), (b) \(12e^{3t}\), (c) \(-12e^{-2t}\)."

Step-by-step solution

Understand the Chain Rule

The chain rule states that for two functions, if we have a composite function \(y = f(g(t))\), the derivative of \(y\) with respect to \(t\) is \(\frac{dy}{dt} = f'(g(t)) \cdot g'(t)\). In this exercise, we will use the chain rule on exponential functions.

Differentiate Part (a)

Given \(y = e^{5t}\), first identify the outer and inner functions. Here, the outer function is \(f(u) = e^u\) and the inner function is \(g(t) = 5t\). The derivative of \(f(u)\) with respect to \(u\) is \(f'(u) = e^u\). The derivative of \(g(t)\) with respect to \(t\) is \(g'(t) = 5\). Applying the chain rule, \(\frac{dy}{dt} = e^{5t} \cdot 5 = 5e^{5t}\).

Differentiate Part (b)

Given \(y = 4e^{3t}\), identify that \(f(u) = 4e^u\) and \(g(t) = 3t\). The constant \(4\) is multiplied outside, so \(f'(u) = 4e^u\). The derivative of \(g(t)\) is \(g'(t) = 3\). Using the chain rule, we find \(\frac{dy}{dt} = 4 \cdot e^{3t} \cdot 3 = 12e^{3t}\).

Differentiate Part (c)

For \(y = 6e^{-2t}\), the functions are \(f(u) = 6e^u\) and \(g(t) = -2t\). The derivative \(f'(u) = 6e^u\) and the derivative \(g'(t) = -2\). By applying the chain rule, the derivative is \(\frac{dy}{dt} = 6 \cdot e^{-2t} \cdot (-2) = -12e^{-2t}\).

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus used to find how a function changes as its input changes. In simpler terms, it helps us determine the rate at which one quantity changes with respect to another. For example, if we have a function that describes the distance a car traveled over time, differentiation allows us to calculate the speed at any given moment.

The process of differentiation involves finding the derivative of a function. The derivative is represented by \frac{dy}{dt}\ when dealing with functions of \(t\). Differentiation is particularly useful when working with curves on a graph. It allows us to find the slope of the tangent line at any point on the curve, giving us insight into the behavior of the function at that specific point.

One essential rule in differentiation, especially when dealing with composite functions, is the chain rule. The chain rule is used when you have a function nested within another function, also known as a function of a function. Understanding how to differentiate using this rule is crucial when you encounter more complex expressions.

Exponential Functions

Exponential functions are a special class of mathematical functions where a constant base is raised to a variable power. They are characterized by rapid growth or decay. The general form of an exponential function is \(f(x) = a \cdot e^{bx}\), where \(e\) is the natural logarithm base, an irrational number approximately equal to 2.71828, and \(a\) and \(b\) are constants.

A unique property of exponential functions is that they are the same as their derivatives, meaning they grow at a rate proportional to their size. This property is frequently used in solving differential equations and modeling real-world scenarios like population growth, radioactive decay, and interest calculations.

In the context of the exercise, exponential functions like \(e^{5t}\), \(4e^{3t}\), and \(6e^{-2t}\) demonstrate how changes to the exponent affect the function's rate of change. By differentiating these functions, we can determine how quickly they increase or decrease as time \(t\) changes.

Composite Functions

Composite functions occur when one function is nested inside another. They can be expressed as \(y = f(g(t))\). Understanding these functions involves recognizing which function acts as the outer function and which one serves as the inner.

To handle composite functions, especially when differentiating, the chain rule becomes indispensable. The chain rule allows us to find the derivative by \frac{dy}{dt} = f'(g(t)) \cdot g'(t)\, meaning we take the derivative of the outer function while keeping the inner function the same, and then multiply it by the derivative of the inner function.

In the given exercise, if we take \(y = e^{5t}\), the outer function \(f(u) = e^u\) and the inner function \(g(t) = 5t\) guide us in differentiating correctly. Similarly, for \(y = 4e^{3t}\) and \(y = 6e^{-2t}\), identifying these inner and outer functions first is a crucial step. Mastering composite functions is not just about being able to differentiate them, but understanding how these layers of functions interact dynamically.