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Chapter 7: Problem 7

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. \begin{equation}y=e^{-5 x}\end{equation}

Short Answer

Expert verified
The derivative is \( \frac{dy}{dx} = -5e^{-5x} \).

Step by step solution

01

Identify the Function to Differentiate

The function given is \( y = e^{-5x} \). We need to find its derivative with respect to \( x \).
02

Apply the Exponential Derivative Rule

Recall that the derivative of \( e^{u} \) with respect to \( x \) is \( e^{u} \times \frac{du}{dx} \). Here, \( u = -5x \), so we need to find \( \frac{du}{dx} \).
03

Differentiate the Exponent

Differentiate \( u = -5x \) with respect to \( x \). The derivative is \( \frac{du}{dx} = -5 \).
04

Use the Chain Rule

Apply the chain rule to differentiate the function. The derivative of \( y = e^{-5x} \) is the original function \( e^{-5x} \) times the derivative of the exponent: \[ \frac{dy}{dx} = e^{-5x} \times (-5) \].
05

Simplify the Derivative

Simplify the expression obtained from the differentiation: \( \frac{dy}{dx} = -5e^{-5x} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Function
An exponential function is a mathematical expression where a constant base is raised to a variable exponent. In the expression \( y = e^{-5x} \), the base \( e \) is the Euler's number, approximately equal to 2.71828. Such functions are powerful because they model growth and decay processes effectively, such as radioactive decay or population growth.

Exponential functions have a unique property: their rate of change is proportional to their value. This leads to their widespread use in real-world applications. In calculus, understanding how these functions change鈥攊.e., finding their derivatives鈥攊s crucial.
Chain Rule
In calculus, the chain rule is a fundamental principle used to find the derivative of a composite function. A composite function is essentially a combination of two or more functions. The rule is essential when dealing with functions nested within each other, like \( y = e^{-5x} \).

Applying the chain rule involves two main steps:
  • Differentiate the outer function, leaving the inner function unchanged.
  • Multiply by the derivative of the inner function.

For our example of \( y = e^{-5x} \), the outer function is \( e^u \) with \( u = -5x \), and its derivative involves exponential and linear differentiation.
Differentiation
Differentiation is the process used to find the derivative of a function. It measures how a function changes as its input changes, and is akin to finding the slope of the tangent to the function's graph at any point.

In determining the derivative of \( y = e^{-5x} \), we used differentiation to find how \( y \) changes with respect to \( x \). This involves understanding how the rate of change of \( e^{-5x} \) depends on \( x \), which is particularly important in calculus for analyzing and interpreting graphs, economic models, and physical phenomena.
Exponential Derivative Rule
The exponential derivative rule provides a straightforward way to differentiate exponential functions. Specifically, for \( e^u \), the rule states that the derivative is \( e^u \times \frac{du}{dx} \). This is derived from the function's unique property where its derivative mirrors its original form.

In our example of \( y = e^{-5x} \), knowing \( u = -5x \), we apply this rule to find that the derivative is \( e^{-5x} \) multiplied by \( -5 \), resulting in \(-5e^{-5x}\). This rule is vital as it simplifies finding derivatives for complex exponential functions, emphasizing the elegance and simplicity in calculus.

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Most popular questions from this chapter

Evaluate the integrals in Exercises \(47-70\) $$ \int \frac{\sec ^{2} y d y}{\sqrt{1-\tan ^{2} y}} $$

In Exercises \(21-42,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\csc ^{-1}\left(e^{t}\right) $$

Evaluate the integrals in Exercises \(85-94\) $$ \int \frac{\sqrt{\tan ^{-1} x} d x}{1+x^{2}} $$

In Exercises \(21-42,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\cos ^{-1}\left(x^{2}\right) $$

Evaluate the integrals in Exercises \(85-94\) $$ \int \frac{d y}{\left(\sin ^{-1} y\right) \sqrt{1-y^{2}}} $$
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