Question

To find the derivative of the function \(G(x) = \frac{{1 - coshx}}{{1 + coshx}}\).

Short answer

The derivative of the function \(G(x) = \frac{{1 - \cosh x}}{{1 + \cosh x}}\) is \(\frac{{ - 2\sinh x}}{{{{(1 + \cosh x)}^2}}}\).

Step-by-step solution

Given function

The function is, \(G(x) = \frac{{1 - \cosh x}}{{1 + \cosh x}}\).

The Concept of derivative

The derivative of\(coshx\)is\(\frac{d}{{dx}}(coshx) = sinhx\).

Evaluate the derivative of given function

Evaluate the derivative of \(G(x) = \frac{{1 - \cosh x}}{{1 + \cosh x}}\).

\(\begin{aligned}{c}{G^\prime }(x) &= {\left( {\frac{{1 - \cosh x}}{{1 + \cosh x}}} \right)^\prime }\\ &= \frac{{(1 + \cosh x){{(1 - \cosh x)}^\prime } - (1 - \cosh x){{(1 + \cosh x)}^\prime }}}{{{{(1 + \cosh x)}^2}}}\\ &= \frac{{(1 + \cosh x)( - \sinh x) - (1 - \cosh x)(\sinh x)}}{{{{(1 + \cosh x)}^2}}}\\ &= \frac{{ - \sinh x - \sinh x\cosh x - \sinh x + \sinh x\cosh x}}{{{{(1 + \cosh x)}^2}}}\end{aligned}\)

So, \( = \frac{{ - 2\sinh x}}{{{{(1 + \cosh x)}^2}}}\).

Therefore, the derivative of the function \(G(x) = \frac{{1 - \cosh x}}{{1 + \cosh x}}\) is \(\frac{{ - 2\sinh x}}{{{{(1 + \cosh x)}^2}}}\).