Question

$\displaystyle y = sec^{-1}(\frac{1+e^{2\sqrt{x}}}{2e^{\sqrt{x}}})$

Answer 1

$\displaystyle \frac{dy}{dx}=\frac{1}{2\sqrt{x}\,\cosh\sqrt{x}}$

Answer 2

We start with

$$y=\sec^{-1}\Bigl(\frac{1+e^{2\sqrt{x}}}{2e^{\sqrt{x}}}\Bigr).$$

Notice that

$$\frac{1+e^{2\sqrt{x}}}{2e^{\sqrt{x}}}=\frac{e^{\sqrt{x}}+e^{-\sqrt{x}}}{2}=\cosh\sqrt{x}.$$

Thus,

$$y=\sec^{-1}\Bigl(\cosh\sqrt{x}\Bigr).$$

The derivative formula for $y=\sec^{-1}(u)$ (with $u>0$) is

$$\frac{dy}{dx}=\frac{u'(x)}{u\,\sqrt{u^2-1}}.$$

Here, $u(x)=\cosh\sqrt{x}$. Differentiate using the chain rule:

$$u'(x)=\frac{d}{dx}\cosh\sqrt{x}=\sinh\sqrt{x}\cdot\frac{1}{2\sqrt{x}}=\frac{\sinh\sqrt{x}}{2\sqrt{x}}.$$

Since

$$\sqrt{u^2-1}=\sqrt{\cosh^2\sqrt{x}-1}=\sqrt{\sinh^2\sqrt{x}}=\sinh\sqrt{x}\quad (\text{for } \sqrt{x}>0),$$

we have

$$\frac{dy}{dx}=\frac{\frac{\sinh\sqrt{x}}{2\sqrt{x}}}{\cosh\sqrt{x}\,\sinh\sqrt{x}}=\frac{1}{2\sqrt{x}\,\cosh\sqrt{x}}.$$