# What is the derivative of this function y=sec^-1(e^(2x))?

Feb 19, 2018

(2)/(sqrt(e^(4x)-1)

#### Explanation:

As if $y = {\sec}^{-} 1 x$ the derivative is equel to $\frac{1}{x \sqrt{{x}^{2} - 1}}$
so by using this formula and if $y = {e}^{2 x}$ then derivative is $2 {e}^{2 x}$ so by using this relation in the formula we get the required answer. as ${e}^{2 x}$ is a function other than $x$ that is why we need further derivative of ${e}^{2 x}$

Feb 19, 2018

$\frac{2}{\sqrt{{e}^{4 x} - 1}}$

#### Explanation:

We have $\frac{d}{\mathrm{dx}} {\sec}^{-} 1 \left({e}^{2 x}\right)$.

We can apply the chain rule, which states that for a function $f \left(u\right)$, its derivative is $\frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

Here, $f = {\sec}^{-} 1 \left(u\right)$, and $u = {e}^{2 x}$.

$\frac{d}{\mathrm{dx}} {\sec}^{-} 1 \left(u\right) = \frac{1}{\sqrt{{u}^{2}} \sqrt{{u}^{2} - 1}}$. This is a common derivative.

$\frac{d}{\mathrm{dx}} {e}^{2 x}$. Chain rule again, here $f = {e}^{u}$ and $x = 2 x$. The derivative of ${e}^{u}$ is ${e}^{u}$, and the derivative of $2 x$ is $2$.

But here, $u = 2 x$, and so we finally have $2 {e}^{2 x}$.

So $\frac{d}{\mathrm{dx}} {e}^{2 x} = 2 {e}^{2 x}$.

Now we have:

$\frac{2 {e}^{2 x}}{\sqrt{{u}^{2}} \sqrt{{u}^{2} - 1}}$, but since $u = {e}^{2 x}$, we have:

$\frac{2 {e}^{2 x}}{\sqrt{{\left({e}^{2 x}\right)}^{2}} \sqrt{{\left({e}^{2 x}\right)}^{2} - 1}}$

$\frac{2 {e}^{2 x}}{{e}^{2 x} \sqrt{\left({e}^{4 x}\right) - 1}}$

$\frac{2}{\sqrt{{e}^{4 x} - 1}}$, our derivative.