Differentiate tangent inverse (cosh inverse of x)?

The derivative of tangent inverse (cosh inverse of x) is given by:

d/dx [tan^(-1)(cosh^(-1)(x))] = 1 / (1 + (cosh^(-1)(x))^2) * d/dx [cosh^(-1)(x)]

To differentiate cosh inverse of x, we can use the chain rule:

d/dx [cosh^(-1)(x)] = -1 / sqrt(1 - x^2) * d/dx [cosh^(-1)(x)]

Therefore, the derivative of tangent inverse (cosh inverse of x) is:

d/dx [tan^(-1)(cosh^(-1)(x))] = -1 / (1 + (cosh^(-1)(x))^2 * sqrt(1 - x^2)

The derivative of the inverse hyperbolic cosine of x, denoted as tanh^(-1)(x), is:

$$\frac{d}{dx} \tanh^{-1}(x) = \frac{1}{1-x^2}$$

To derive this, you can use the following steps:

1. Let y = tanh^(-1)(x).

2. Then, tanh(y) = x.

3. Differentiate both sides of the equation with respect to x using the chain rule:

```
sech^2(y) dy/dx = 1
```

4. Solve the equation for dy/dx:

```
dy/dx = 1/sech^2(y)
```

5. Substitute the identity sech^2(y) = 1 - tanh^2(y) into the equation:

```
dy/dx = 1/(1 - tanh^2(y))
```

6. Substitute y = tanh^(-1)(x) into the equation:

```
dy/dx = 1/(1 - x^2)
```

Therefore, the derivative of tanh^(-1)(x) is 1/(1-x^2).