Question

How do you find the derivative of \[h\left( x \right) = \ln \left( {\cosh \left( {2x} \right)} \right)\] ?

Answer 1

Hint: Given is a hyperbolic function. To find the derivative we will use a method of substitution and chain rule for the solution because the function is a composite function that has two different functions . We will consider \[u = \cosh \left( {2x} \right)\] . Then using the chain rule \[\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}\] , we will find the derivative. We will substitute the value of u. This will be the way to solve it.

Complete step-by-step answer:
Given that,
\[h\left( x \right) = \ln \left( {\cosh \left( {2x} \right)} \right)\]
Consider, \[y = h\left( x \right) = \ln \left( {\cosh \left( {2x} \right)} \right)\]
We know that, product rule is written as,
\[\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}\]
We will find the two factors of the product.
Putting \[u = \cosh \left( {2x} \right)\]
\[\dfrac{d}{{dx}}\cosh \left( {2x} \right) = 2\sinh \left( {2x} \right)\]
\[\dfrac{{du}}{{dx}} = 2\sinh \left( {2x} \right)\]
Putting the value of u,
\[y = \ln u\]
Finding the derivative,
\[\dfrac{{dy}}{{du}} = \dfrac{1}{u}\]
Putting the values in the chain rule;
\[\dfrac{{dy}}{{dx}} = \dfrac{1}{u} \times 2\sinh \left( {2x} \right)\]
Substitute the value of u,
\[\dfrac{{dy}}{{dx}} = \dfrac{{2\sinh \left( {2x} \right)}}{{\cosh \left( {2x} \right)}}\]
Ratio of sin to cos is tan,
\[\dfrac{{dy}}{{dx}} = 2\tanh \left( {2x} \right)\]
This is the correct answer.
So, the correct answer is “\[\dfrac{{dy}}{{dx}} = 2\tanh \left( {2x} \right)\] ”.

Note: Note that the given function is hyperbolic function. This given function is a combination of two such functions. So we have used chain rule. In that, we have two derivatives. Hyperbolic functions are the trigonometric functions that use a hyperbola to define the function rather than using a circle.