Question

find the derivative of y= log (cosh 2x)

Answer 1

y = Log[Cosh[2 x]]

dy/dx = (1/Cosh[2x])*(d/dx(Cosh[2x]))

= (1/Cosh[2x])*(2 Sinh[2x])

=2 Tanh[2x]

Answer 2

The derivative of the given function $y=\log( \cosh 2x)$ is $2\tanh 2x$.

Given,

Function:

$y=\log( \cosh 2x).$

To find,

Derivative of the given function.

Solution,

The derivative of a function, in mathematics, is defined as the rate of change of the function with respect to the variable on which the function depends.

Here, the given function is,

$y=\log( \cosh 2x)$.

If a function is such that it is dependent on any other function also. Then its derivative is determined as follows.

Let the function is

$y=f(f(x))$

then,

$\frac{dy}{dx} =f'(f(x)) \times f'(x)$

where,

$f'$ represents derivative or differentiation of function $f$.

Now, for the given function, we can find the derivative as follows.

Given function $y=\log( \cosh 2x)$.

$\frac{dy}{dx}= \frac{d}{dx} (\log( \cosh 2x)) \times \frac{d}{dx} ( \cosh 2x) \times \frac{d}{dx} (2x)$

We know that,

$\frac{d}{dx} (\log x)=\frac{1}{x},$

$\frac{d}{dx} (\cosh x)=\sinh x,$ and

$\frac{d}{dx} (2 x)=2.$

Thus,

$\frac{dy}{dx}= \frac{1}{\cosh 2x} \times ( \sinh 2x) \times (2)$

Simplifying,

$\frac{dy}{dx}= \frac{2\sinh 2x}{\cosh 2x}$

$\implies \frac{dy}{dx}= 2\tanh 2x$

Therefore, the derivative of the given function $y=\log( \cosh 2x)$ is $2\tanh 2x$.

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