Question

Differentiate In(coshx-1/coshx+1)

Answer 1

To differentiate the function \( y = \frac{\cosh x - 1}{\cosh x + 1} \), we can apply the quotient rule. The quotient rule states that if \( y = \frac{u}{v} \), then

\[
\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}
\]

In this case, let:
- \( u = \cosh x - 1 \)
- \( v = \cosh x + 1 \)

First, we'll compute the derivatives of \( u \) and \( v \):

\[
\frac{du}{dx} = \sinh x
\]
\[
\frac{dv}{dx} = \sinh x
\]

Now, using the quotient rule:

\[
\frac{dy}{dx} = \frac{(\cosh x + 1)(\sinh x) - (\cosh x - 1)(\sinh x)}{(\cosh x + 1)^2}
\]

Now, simplify the numerator:

\[
= \frac{\sinh x (\cosh x + 1 - \cosh x + 1)}{(\cosh x + 1)^2}
\]
\[
= \frac{\sinh x (2)}{(\cosh x + 1)^2}
\]
\[
= \frac{2 \sinh x}{(\cosh x + 1)^2}
\]

Thus, the derivative of \( y \) with respect to \( x \) is:

\[
\frac{dy}{dx} = \frac{2 \sinh x}{(\cosh x + 1)^2}
\]