Chapter 3: Problem 51
Calculate the derivative of the following functions. $$\sqrt{x+\sqrt{x}}$$
Short Answer
Expert verified
Answer: The derivative of the function \(y = \frac{2e^x + 3e^{-x}}{3}\) with respect to \(x\) is \(\frac{dy}{dx} = \frac{2e^x - 3e^{-x}}{3}\).
Step by step solution
01
Identify the function components
We start by identifying each component of the function:
$$y=\frac{2 e^{x}+3 e^{-x}}{3}$$
Here, we have the sum of two functions: \(2e^x\) and \(3e^{-x}\). We will differentiate each of them separately and then divide by 3.
02
Differentiate \(2e^x\)
Using the rules mentioned above, we differentiate the first part, \(2e^x\), with respect to \(x\):
$$\frac{d}{dx}(2e^{x}) = 2\frac{d}{dx}(e^{x}) = 2e^{x}$$
03
Differentiate \(3e^{-x}\)
Now, we differentiate the second part, \(3e^{-x}\), with respect to \(x\):
$$\frac{d}{dx}(3e^{-x}) = 3\frac{d}{dx}(e^{-x})$$
Applying the chain rule (derivative of the outer function times the derivative of the inner function), we get:
$$3\frac{d}{dx}(e^{-x}) = 3(-e^{-x})$$
04
Combine the derivative of the two components
Now that we have the derivatives of \(2e^x\) and \(3e^{-x}\), we can combine them to find the overall derivative of the function:
$$\frac{d}{dx}\left(\frac{2 e^{x}+3 e^{-x}}{3}\right) = \frac{1}{3}\left(\frac{d}{dx}(2 e^{x})+\frac{d}{dx}(3 e^{-x})\right)$$
Substitute the derivatives we found in Steps 2 and 3:
$$\frac{d}{dx}\left(\frac{2 e^{x}+3 e^{-x}}{3}\right) = \frac{1}{3}(2e^{x} - 3e^{-x})$$
The derivative of the function, \(y=\frac{2e^x+3e^{-x}}{3}\), with respect to \(x\) is:
$$\frac{dy}{dx} = \frac{2e^x - 3e^{-x}}{3}$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Chain Rule
The **chain rule** is an essential tool in calculus that helps us differentiate composite functions. A composite function is simply a function within another function. Let's consider a function structured as \((f(g(x)))\). To differentiate such a function, we apply the chain rule. This involves differentiating the outer function, \(f\), first, as if \(g(x)\) is a standalone, then multiplying the result by the derivative of the inner function, \(g(x)\).
The formula for the chain rule is articulated as: \[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]
The formula for the chain rule is articulated as: \[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]
- Outer function \(f(u)\) where \(u = g(x)\)
- Inner function \(g(x)\)
- Derivative of Inner \(g'(x)\)
Insights into Exponential Functions
**Exponential functions** are common building blocks in calculus and come in the form \(e^{x}\), where \(e\) is Euler's number, approximately equal to 2.71828. What makes exponential functions unique is their growth rate and straightforward differentiation process.
For any exponential function \(e^{f(x)}\), the derivative is determined by employing both the chain rule and understanding the basics of derivative operations on exponentials. Specifically, the derivative of \(e^{x}\) is \(e^{x}\). If we modify the exponent to any function \(f(x)\), the chain rule helps us differentiate by tackling the exponent's derivative.
In the example provided, the expressions \(2e^{x}\) and \(3e^{-x}\) illustrate how exponential functions are differentiated. For \(2e^{x}\), we took the derivative as \(2 imes e^{x}\). Meanwhile, \(3e^{-x}\) required applying the chain rule to account for the modification in the exponent, producing \(3(-e^{-x})\). Understanding these operations prepares us to handle more complex scenarios involving exponential growth or decay.
For any exponential function \(e^{f(x)}\), the derivative is determined by employing both the chain rule and understanding the basics of derivative operations on exponentials. Specifically, the derivative of \(e^{x}\) is \(e^{x}\). If we modify the exponent to any function \(f(x)\), the chain rule helps us differentiate by tackling the exponent's derivative.
In the example provided, the expressions \(2e^{x}\) and \(3e^{-x}\) illustrate how exponential functions are differentiated. For \(2e^{x}\), we took the derivative as \(2 imes e^{x}\). Meanwhile, \(3e^{-x}\) required applying the chain rule to account for the modification in the exponent, producing \(3(-e^{-x})\). Understanding these operations prepares us to handle more complex scenarios involving exponential growth or decay.
Differentiation Rules Recap
**Differentiation rules** are the foundation of finding derivatives of various types of functions. These rules provide the steps and processes to determine the derivative of a function efficiently. Several important rules include:
- Power Rule: For \(x^n\), the derivative is \(nx^{n-1}\).
- Constant Rule: The derivative of a constant is zero.
- Sum Rule: The derivative of a sum of functions is the sum of the derivatives.
- Product Rule: For products of functions \(u(x)v(x)\), it is \(u'v + uv'\).
- Quotient Rule: For \(\frac{u(x)}{v(x)}\), it is \(\frac{u'v - uv'}{v^2}\).