Let y(x)=ln(cosh(x)+ sqrt(cosh^2(x)−1))

Find the derivative Dxy=

I don't know what to do with this...

To find the derivative of y(x) = ln(cosh(x) + sqrt(cosh^2(x) - 1), you can use the chain rule.

Step 1: Identify the inner function and the outer function.
The inner function is cosh(x) + sqrt(cosh^2(x) - 1), and the outer function is ln(x).

Step 2: Find the derivative of the outer function.
The derivative of ln(x) is 1/x.

Step 3: Find the derivative of the inner function.
For the inner function, we need to use the chain rule. Let's denote the inner function as u(x).
Let u(x) = cosh(x) + sqrt(cosh^2(x) - 1).

Now, let's find the derivative of u(x):
du(x)/dx = d(cosh(x) + sqrt(cosh^2(x) - 1))/dx.

To find the derivative, we need to find the derivative of each term separately and then apply the sum rule.

The derivative of cosh(x) is sinh(x).
The derivative of sqrt(cosh^2(x) - 1) can be found by applying the chain rule.
Let v(x) = cosh^2(x) - 1.
Then, sqrt(v(x)) = (v(x))^(1/2).
Using the chain rule, the derivative of sqrt(v(x)) is (1/2) * (v(x))^(-1/2) * dv(x)/dx.
Now, let's find dv(x)/dx:
dv(x)/dx = d(cosh^2(x) - 1)/dx = 2 * cosh(x) * sinh(x) - 0 = 2 * cosh(x) * sinh(x).

Plugging these into the derivative of u(x):
du(x)/dx = sinh(x) + (1/2) * (cosh^2(x) - 1)^(-1/2) * 2 * cosh(x) * sinh(x).

Step 4: Apply the chain rule to find the derivative of y(x).
Using the chain rule, the derivative of y(x) = ln(u(x)) is given by:

Dxy = (dy/dx) = (du(x)/dx) * (1/u(x)).

Now, let's substitute the derivatives we found earlier:
Dxy = (sinh(x) + (1/2) * (cosh^2(x) - 1)^(-1/2) * 2 * cosh(x) * sinh(x)) / (cosh(x) + sqrt(cosh^2(x) - 1)).

Thus, the derivative of y(x) = ln(cosh(x) + sqrt(cosh^2(x) - 1)) is Dxy = (sinh(x) + (1/2) * (cosh^2(x) - 1)^(-1/2) * 2 * cosh(x) * sinh(x)) / (cosh(x) + sqrt(cosh^2(x) - 1)).

To find the derivative of the function y(x) = ln(cosh(x) + sqrt(cosh^2(x)−1)), we can use the chain rule. The chain rule states that if we have a function of the form f(g(x)), the derivative can be found by taking the derivative of the outer function f'(g(x)) and multiplying it by the derivative of the inner function g'(x).

Let's break down the given function to make it easier to differentiate.

y(x) = ln(cosh(x) + sqrt(cosh^2(x)−1))

Let's define two inner functions:
u(x) = cosh(x)
v(x) = sqrt(cosh^2(x)−1)

Now we can rewrite the function as:
y(x) = ln(u(x) + v(x))

Applying the chain rule, we have:
Dxy = (dy/du) * (du/dx) + (dy/dv) * (dv/dx)

Let's find the derivatives of u(x) and v(x):

du/dx = sinh(x) (derivative of hyperbolic cosine)

dv/dx = (1/2) * (cosh^2(x)−1)^(-1/2) * 2 * cosh(x) * sinh(x)
= sinh(x) / sqrt(cosh^2(x)−1)

Now let's find the derivatives of y(x) with respect to u(x) and v(x):

(dy/du) = 1 / (u(x) + v(x)) (derivative of natural logarithm)

(dy/dv) = 1 / (u(x) + v(x)) (derivative of natural logarithm)

Finally, we can substitute all the derivatives into the equation Dxy = (dy/du) * (du/dx) + (dy/dv) * (dv/dx):

Dxy = ((1 / (u(x) + v(x))) * sinh(x)) + ((1 / (u(x) + v(x))) * (sinh(x) / sqrt(cosh^2(x)−1)))

Simplifying further, we get:

Dxy = sinh(x) * ((1 / (u(x) + v(x))) + (1 / ((u(x) + v(x)) * sqrt(cosh^2(x)−1)))

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

Dxy = sinh(x) * (1 + (1 / sqrt(cosh^2(x)−1))) / (cosh(x) + sqrt(cosh^2(x)−1))