Derivative of acosh(x) - Proof and Explanation

Proof

1. Definition of $\text{acosh}(x)$:

   $$y=\text{acosh}(x)⟹x=\cosh (y)$$

   where $\cosh (y)=\frac{e^y+e^{-y}}{2}$.

2. Differentiate both sides with respect to $x$:

   $$\frac{\text{d}}{\text{d}x}x=\frac{\text{d}}{\text{d}x}\cosh (y)$$

3. Using the chain rule on the right side:

   $$1=\sinh (y)·\frac{dy}{dx}$$

   where $\sinh (y)=\frac{e^y-e^{-y}}{2}$.

4. Solve for $\frac{dy}{dx}$:

   $$\frac{dy}{dx}=\frac{1}{\sinh (y)}$$

5. Express $\sinh (y)$ in terms of $x$:

   $${\cosh }^2(y)-1={\sinh }^2(y)$$

   Since $x=\cosh (y)$,

   $${\sinh }^2(y)=x^2-1⟹\sinh (y)=\sqrt{x^2-1}$$

6. Substitute back into the derivative:

   $$\frac{dy}{dx}=\frac{1}{\sqrt{x^2-1}}$$

Thus, the derivative of $\text{acosh}(x)$ is:

Explanation

To understand the derivative of $\text{acosh}(x)$, we first need to understand the function itself. The inverse hyperbolic cosine function, $\text{acosh}(x)$, is defined as the inverse of the hyperbolic cosine function, $\cosh (y)$. This means if $y=\text{acosh}(x)$, then $x=\cosh (y)$, where $\cosh (y)=\frac{e^y+e^{-y}}{2}$.

To find the derivative of $\text{acosh}(x)$, we start by differentiating both sides of the equation $x=\cosh (y)$ with respect to $x$. This gives us:

Using the chain rule, the derivative of $\cosh (y)$ with respect to $x$ is $\sinh (y)·\frac{dy}{dx}$, where $\sinh (y)=\frac{e^y-e^{-y}}{2}$. So we have:

Next, we solve for $\frac{dy}{dx}$:

To express $\sinh (y)$ in terms of $x$, we use the identity ${\cosh }^2(y)-1={\sinh }^2(y)$. Since $x=\cosh (y)$, we substitute $\cosh (y)$ with $x$ to get:

Taking the square root, we find:

Finally, we substitute $\sinh (y)$ back into the derivative:

Therefore, the derivative of $\text{acosh}(x)$ with respect to $x$ is $\boxed{\frac{1}{\sqrt{x^2-1}}}$.

Q.E.D.