# How do you differentiate f(x)=lnsqrt(-e^(4x)-2) using the chain rule.?

Mar 13, 2018

$\frac{2 {e}^{4 x}}{{e}^{4 x} + 2}$

#### Explanation:

According to the chain rule, $\left(f \left(g \left(x\right)\right)\right) ' = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$.

We have $f \left(u\right) = \ln u$ where $u = g \left(x\right) = \sqrt{- {e}^{4 x} - 2}$

We need to find $\frac{d}{\mathrm{du}} \ln u \cdot \frac{d}{\mathrm{dx}} \sqrt{- {e}^{4 x} - 2}$

Since $\frac{d}{\mathrm{dx}} \ln x = \frac{1}{x}$, we write:

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

Concentrating on the bracket, let's define two other functions:

$h \left(v\right) = \sqrt{v}$ where $v = j \left(x\right) = - {e}^{4 x} - 2$

Since $\frac{d}{\mathrm{dx}} \sqrt{x} = \frac{1}{2 \sqrt{x}}$, we write:

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

Concentrating on the bracket, we define two more functions:

$k \left(z\right) = - {e}^{z} - 2$, where $z = l \left(x\right) = 4 x$

We have $\frac{d}{\mathrm{dz}} \left(- {e}^{z} - 2\right) \cdot \frac{d}{\mathrm{dx}} \left(4 x\right)$

$- \left(\frac{d}{\mathrm{dz}} {e}^{z} + \frac{d}{\mathrm{dz}} 2\right) \cdot 4 \frac{d}{\mathrm{dx}} x$

$- \left({e}^{z}\right) \cdot 4$

$- 4 {e}^{z}$

Overall, we have:

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

Well, that's not right, right? We just have a bunch of nonsense variables. But remember that:

$u = \sqrt{- {e}^{4 x} - 2}$

$v = - {e}^{4 x} - 2$

$z = 4 x$

We can input our stuff in:

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

And simplify:

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

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

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

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

$\frac{2 {e}^{4 x}}{{e}^{4 x} + 2}$