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Chapter 3: Problem 22

Find the following derivatives. $$\frac{d}{d x}\left(\ln \left(e^{x}+e^{-x}\right)\right)$$

Short Answer

Expert verified
Question: Find the derivative of the function $$\ln\left(e^x + e^{-x}\right)$$ with respect to \(x\). Solution: The derivative of the function is given by: $$\frac{d}{dx}\left(\ln\left(e^x + e^{-x}\right)\right) = \frac{e^x - e^{-x}}{e^x + e^{-x}}$$.

Step by step solution

01

Identify the Chain Rule

We need to find the derivative of $$\ln\left(e^{x}+e^{-x}\right)$$ with respect to \(x\). Notice that this function is in the form of $$\ln(u)$$, where $$u = e^x + e^{-x}$$. We can apply the chain rule to find the derivative, which states that if you have a composite function $$y = f(g(x))$$, then the derivative of $$y$$ with respect to $$x$$ is: $$\frac{dy}{dx} = \frac{dy}{du} * \frac{du}{dx}$$ In this case, our function is $$y = \ln(u)$$, so: $$\frac{dy}{dx} = \frac{d\ln(u)}{du} * \frac{du}{dx}$$ Now, we'll find the two derivatives: $$\frac{d\ln(u)}{du}$$ and $$\frac{du}{dx}$$.
02

Calculate the Derivative of Natural Log Function

First, let's find the derivative of the natural logarithm function with respect to u, which is $$\frac{d\ln(u)}{du}$$. The derivative of the natural logarithm function is simply the reciprocal of the argument: $$\frac{d\ln(u)}{du} = \frac{1}{u}$$ Substitute our expression for $$u$$: $$\frac{d\ln(u)}{du} = \frac{1}{e^x + e^{-x}}$$
03

Calculate the Derivative of the Argument Function

Now, let's find the derivative of the argument function with respect to $$x$$, which is $$\frac{du}{dx}$$, where $$u = e^x + e^{-x}$$. By applying the sum rule, we can find the derivative of each term separately: $$\frac{du}{dx} = \frac{d(e^x)}{dx} + \frac{d(e^{-x})}{dx}$$ The derivative of $$e^x$$ with respect to $$x$$ is simply $$e^x$$. For the second term, we'll need to apply the chain rule since it's a composite function. Let $$g(x) = -x$$, then $$e^{-x} = e^{g(x)}$$. Applying the chain rule: $$\frac{d(e^{-x})}{dx} = \frac{d(e^{g(x)})}{dg} * \frac{dg}{dx} = e^{g(x)} * (-1) = -e^{-x}$$ Now we can substitute our derivatives back into the expression: $$\frac{du}{dx} = e^{x} - e^{-x}$$
04

Use the Chain Rule to Find the Derivative of the Function

Now that we have expressions for both $$\frac{d\ln(u)}{du}$$ and $$\frac{du}{dx}$$, we can plug them back into the chain rule equation: $$\frac{d\ln\left(e^x + e^{-x}\right)}{dx} = \frac{d\ln(u)}{du} * \frac{du}{dx} = \frac{1}{e^x + e^{-x}} * (e^x - e^{-x})$$ So, the derivative of the given function is: $$\frac{d}{dx}\left(\ln\left(e^x + e^{-x}\right)\right) = \frac{e^x - e^{-x}}{e^x + e^{-x}}$$

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