Question

differentiate ln(e^x + e^-x) wrt x

Answer 1

To differentiate the function \( y = \ln(e^x + e^{-x}) \) with respect to \( x \), we can apply the chain rule and the properties of logarithms.

1. Start with the function:
\[
y = \ln(e^x + e^{-x})
\]

2. Differentiate using the chain rule:
\[
\frac{dy}{dx} = \frac{1}{e^x + e^{-x}} \cdot \frac{d}{dx}(e^x + e^{-x})
\]

3. Now, differentiate \( e^x + e^{-x} \):
\[
\frac{d}{dx}(e^x + e^{-x}) = e^x - e^{-x}
\]

4. Substitute this back into the derivative:
\[
\frac{dy}{dx} = \frac{1}{e^x + e^{-x}} \cdot (e^x - e^{-x})
\]

5. So the final result for the derivative is:
\[
\frac{dy}{dx} = \frac{e^x - e^{-x}}{e^x + e^{-x}}
\]

Thus, the derivative of \( \ln(e^x + e^{-x}) \) with respect to \( x \) is:
\[
\frac{dy}{dx} = \frac{e^x - e^{-x}}{e^x + e^{-x}}
\]