Question Statement

Find the derivative $\frac{dy}{dx}$ for the function:

$y=\ln \left(e^x+e^{-x}\right)$

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Background and Explanation

This problem requires differentiating a logarithmic function with a composite argument involving exponential terms. You should be familiar with the chain rule and the derivatives of exponential functions before attempting this problem.

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Solution

We are given the function $y=\ln \left(e^x+e^{-x}\right)$ and need to find its derivative with respect to $x$.

This is a composite function where the outer function is the natural logarithm $\ln (u)$ and the inner function is $u=e^x+e^{-x}$. To differentiate this, we apply the chain rule.

Step 1: Apply the chain rule for logarithmic differentiation.

The derivative of $\ln (u)$ with respect to $x$ is $\frac{1}{u}\cdot \frac{du}{dx}$. Substituting our inner function:

$\frac{dy}{dx}=\frac{1}{e^x+e^{-x}}\cdot \frac{d}{dx}\left(e^x+e^{-x}\right)$

Step 2: Differentiate the inner function.

We differentiate term by term:

• The derivative of $e^x$ is $e^x$
• The derivative of $e^{-x}$ requires the chain rule: $\frac{d}{dx}(e^{-x})=e^{-x}\cdot (-1)=-e^{-x}$

Therefore: $\frac{d}{dx}\left(e^x+e^{-x}\right)=e^x-e^{-x}$

Step 3: Substitute and simplify.

Substituting the result from Step 2 back into our expression from Step 1:

$\begin{array}{rl}\frac{dy}{dx}& =\frac{1}{e^x+e^{-x}}\cdot \left(e^x-e^{-x}\right)\\ & =\frac{e^x-e^{-x}}{e^x+e^{-x}}\end{array}$

Step 4: Recognize the hyperbolic tangent.

The resulting expression matches the definition of the hyperbolic tangent function:

$\tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}$

Thus, we can write the final answer as:

$\frac{dy}{dx}=\tanh x$

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Key Formulas or Methods Used

• Chain Rule for Logarithms: $\frac{d}{dx}[\ln (u)]=\frac{1}{u}\cdot \frac{du}{dx}$
• Derivative of Exponential: $\frac{d}{dx}(e^x)=e^x$ and $\frac{d}{dx}(e^{-x})=-e^{-x}$
• Definition of Hyperbolic Tangent: $\tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}$

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Summary of Steps

1. Identify the composite structure: outer function $\ln (u)$ where $u=e^x+e^{-x}$
2. Apply the chain rule: $\frac{dy}{dx}=\frac{1}{e^x+e^{-x}}\cdot \frac{d}{dx}(e^x+e^{-x})$
3. Differentiate the inner function to get $e^x-e^{-x}$
4. Multiply to obtain $\frac{e^x-e^{-x}}{e^x+e^{-x}}$
5. Recognize and simplify to $\tanh x$ using the definition of hyperbolic tangent