Question

Find the derivative of the function. \(y=\ln \left(\frac{1+e^{x}}{1-e^{x}}\right)\)

Short answer

The derivative of the function \(y=\ln \left(\frac{1+e^{x}}{1-e^{x}}\right)\) is -e^x.

Step-by-step solution

Identification of Functions

Identify the outer function as \(y = \ln u \) and inner function as \(u = \frac{1+e^x}{1-e^x} \).

Derivation of the Outer Function

Derive the outer function. The derivative of y with respect to u is \(\frac{d}{du}(y)=1/u \). Note that the derivative of the natural logarithm, ln(u), is 1/u.

Derivation of the Inner Function

Derive the inner function using the quotient rule. The derivative of u with respect to x is \(\frac{d}{dx}(u)= \frac{(1-e^x)(1)-(1+e^x)(-e^x)}{(1-e^x)^2} = \frac{e^x}{(1-e^x)^2} \). The quotient rule states that the derivative of \(v/w\) is \((wv' - vw')/w^2 \), where v' and w' are the derivatives of v and w respectively.

Apply the Chain Rule

Multiply the derivatives from Step 2 and 3, this is the chain rule. The chain rule states that \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). This gives, \(\frac{dy}{dx} = \frac{1}{u} \cdot \frac{e^x}{(1-e^x)^2} = \frac{e^x}{(1 + e^x)(1 - e^x)}\).

Simplify the Function

Now, reintroduce \(u = \frac{1+e^x}{1-e^x} \) into the function from step 4. This gives, \(\frac{dy}{dx} = \frac{e^x}{u} = \frac{e^x}{\frac{1+e^x}{1-e^x}} \). Simplifying yields \(\frac{dy}{dx} = \frac{e^x (1-e^x)}{1+e^x} = -e^x\).