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

Find the derivative Dxy=

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

We want to find the derivative of y(x) = ln(cosh(x) + sqrt(cosh^2(x) - 1)) with respect to x. To do this, we'll use the chain rule, which states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. So:

D[y(x)] = D[ln(u)] * D[u]

Where u = cosh(x) + sqrt(cosh^2(x) - 1).

First, let's find the derivative of the outer function, ln(u). The derivative of the natural logarithm is 1/u:

D[ln(u)] = 1/u.

Now, let's find the derivative of the inner function, u = cosh(x) + sqrt(cosh^2(x) - 1). We'll need to use the chain rule again, as well as the derivative of the hyperbolic cosine function, which is the hyperbolic sine function (sinh(x)):

D[u] = D[cosh(x)] + D[sqrt(cosh^2(x) - 1)]

D[u] = sinh(x) + (1/2)(cosh^2(x) - 1)^(-1/2) * D[cosh^2(x) - 1]

D[u] = sinh(x) + (1/2)(cosh^2(x) - 1)^(-1/2) * (2*cosh(x)*sinh(x) - 0)

D[u] = sinh(x) + (cosh(x)*sinh(x))/(sqrt(cosh^2(x) - 1))

Now that we have both the derivatives of the outer and inner functions, we can apply the chain rule:

D[y(x)] = D[ln(u)] * D[u] = (1/u) * D[u].

Substituting the expressions for the inner function (u) and its derivative (D[u]):

D[y(x)] = 1/(cosh(x) + sqrt(cosh^2(x) - 1)) * (sinh(x) + (cosh(x)*sinh(x))/(sqrt(cosh^2(x) - 1)))

This is the derivative of y(x) with respect to x.

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

Let's break down the function to make it easier to work with:

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

Now, let's find the derivative of u with respect to x using the chain rule. The chain rule states that if y = f(u) and u = g(x), then the derivative of y with respect to x is given by dy/dx = f'(u) * g'(x).

In this case:
f(u) = ln(u)
g(x) = cosh(x) + sqrt(cosh^2(x)−1)

First, find the derivative of f(u):
f'(u) = 1/u

Next, find the derivative of g(x):

The derivative of cosh(x) is sinh(x).
The derivative of sqrt(cosh^2(x)−1) can be found using the chain rule. Let s = cosh(x)−1, then sqrt(s) = s^1/2. The derivative of sqrt(s) is (1/2)s^(-1/2) (ds/dx).
So, ds/dx = sinh(x), and therefore, the derivative of sqrt(cosh^2(x)−1) is (1/2)(cosh(x)−1)^(-1/2)sinh(x).

Now, let's combine the derivatives from the chain rule:

dy/dx = f'(u) * g'(x)
= (1/u) * (sinh(x) + (1/2)(cosh(x)−1)^(-1/2)sinh(x))

Substituting u back in, we get:

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

This is the derivative of the given function y(x).

To find the derivative of the given function, we can use the chain rule.

The chain rule states that if we have a composite function y = f(g(x)), where f(u) and g(x) are differentiable functions, then the derivative of y with respect to x is given by:

dy/dx = f'(g(x)) * g'(x)

In our case, let's consider the function y(x) = ln(cosh(x) + sqrt(cosh^2(x) - 1)):

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

Step 2: Find the derivative of the outer and inner functions.
- The derivative of the natural logarithm function ln(u) is given by 1/u times the derivative of u with respect to x.
So, the derivative of the outer function is 1/u * du/dx.

- The derivative of the inner function u(x) = cosh(x) + sqrt(cosh^2(x) - 1) requires a bit more work.
Let's denote u(x) = f(x) + g(x), where f(x) = cosh(x) and g(x) = sqrt(cosh^2(x) - 1).
The derivative of the sum of functions is the sum of the derivatives of each function.
So, the derivative of u(x) is given by du/dx = df/dx + dg/dx.

Now, let's find the derivatives of f(x) = cosh(x) and g(x) = sqrt(cosh^2(x) - 1):
- The derivative of f(x) = cosh(x) is sinh(x).
- The derivative of g(x) requires the chain rule. Let's denote h(x) = cosh^2(x) - 1.
Then, g(x) = sqrt(h(x)). The derivative of g(x) is given by dg/dx = dh/dx * (1/2)h^(-1/2).
Let's calculate dh/dx:
- The derivative of h(x) = cosh^2(x) - 1 is given by the derivative of cosh^2(x) minus the derivative of 1.
- The derivative of cosh^2(x) is 2cosh(x) * sinh(x).
- The derivative of 1 is 0.
So, dh/dx = 2cosh(x) * sinh(x).

Therefore, the derivative of g(x) = sqrt(cosh^2(x) - 1) is given by dg/dx = (1/2)(2cosh(x) * sinh(x)) * (cosh^2(x) - 1)^(-1/2).

Step 3: Put it all together.
- Now that we have the derivative of the outer function (1/u * du/dx) and the derivative of the inner function (df/dx + dg/dx), we can substitute them into the chain rule formula.

dy/dx = (1/u * du/dx) = (1/(cosh(x) + sqrt(cosh^2(x) - 1))) * (du/dx)
= (1/(cosh(x) + sqrt(cosh^2(x) - 1))) * (df/dx + dg/dx)
= (1/(cosh(x) + sqrt(cosh^2(x) - 1))) * (sinh(x) + (1/2)(2cosh(x) * sinh(x)) * (cosh^2(x) - 1)^(-1/2))

This expression represents the derivative of the function y(x) with respect to x.