Chapter 3: Problem 5
In Exercises \(1-57,\) find the derivatives. Assume that \(a, b,\) and \(c\) are constants. $$y=\sqrt{e^{x}+1}$$
Short Answer
Expert verified
The derivative is \( y' = \frac{e^x}{2\sqrt{e^x + 1}}.\)
Step by step solution
01
Recognize the Function Type
The function given is a composite function consisting of an outer function, which is a square root, and an inner function, which is an exponential function plus a constant: \( y = \sqrt{e^x + 1} \).
02
Apply the Chain Rule
To differentiate a composite function, use the chain rule. The chain rule states that the derivative of \( y = f(g(x)) \) is \( y' = f'(g(x)) \cdot g'(x) \). Here, \( f(u) = \sqrt{u} \) and \( g(x) = e^x + 1 \).
03
Differentiate the Outer Function
Differentiate the outer function, \( f(u) = \sqrt{u} \). The derivative is \( f'(u) = \frac{1}{2\sqrt{u}} \) because the derivative of \( \sqrt{u} \) is \( \frac{1}{2} u^{-1/2} \).
04
Differentiate the Inner Function
Differentiate the inner function, \( g(x) = e^x + 1 \). The derivative is \( g'(x) = e^x \) since the derivative of \( e^x \) is \( e^x \), and the derivative of a constant is 0.
05
Combine the Derivatives
Substitute \( u = e^x + 1 \), \( f'(u) = \frac{1}{2\sqrt{u}} \), and \( g'(x) = e^x \) back into the chain rule: \[ y' = \frac{1}{2\sqrt{u}} \cdot g'(x) = \frac{1}{2\sqrt{e^x + 1}} \cdot e^x. \]
06
Simplify the Expression
The expression simplifies to: \[ y' = \frac{e^x}{2\sqrt{e^x + 1}}. \] This is the final form of the derivative.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
The chain rule is a powerful technique in calculus used to differentiate composite functions. It comes into play when we have a function inside of another function, like a nested doll. To use the chain rule, recognize your function as a composition of two simpler functions. Suppose you have a function of the form
- \( y = f(g(x)) \).
- \( y' = f'(g(x)) \cdot g'(x) \).
Composite Functions
Composite functions involve plugging one function into another. Think of them as two-step machines: first, an input goes through the first function, and the result flows into the second function. A composite function is written as:
- \( y = f(g(x)) \).
Differentiation
Differentiation is the process of finding a derivative, a central concept in calculus that represents the rate of change of a function. It answers the question: given a function, how does it change as its input changes? Here's the step-by-step:
- Identify the function you want to differentiate.
- Apply the appropriate differentiation rules, such as the power rule, product rule, or chain rule.
- Simplify the derivative, if possible.
Exponential Functions
Exponential functions involve expressions where the variable appears in the exponent, such as \( e^x \). These functions describe processes that grow or decay at constant rates, prevalent in areas like finance, biology, and physics. Important characteristics of exponential functions include:
- An exponential function can be written as \( a \cdot b^x \), with \( e^x \) being a special case where \( b \) is the base \( e \), approximately equal to 2.718.
- They have a unique property: the rate of change of the function is proportional to itself, which makes them very predictable.
- The derivative of \( e^x \) with respect to \( x \) is \( e^x \).