# How do you differentiate f(x) = ln(sqrt(arcsin(e^(3x)) )  using the chain rule?

Jan 23, 2016

$f ' \left(x\right) = \frac{3 {e}^{3 x}}{2 \sqrt{1 - {e}^{6}} \arcsin \left({e}^{3 x}\right)}$

#### Explanation:

The first step we should take is to simplify the function by using logarithm rules to remove the square root from the function.

Since the square root is really a $1 \text{/} 2$ power, and $\ln \left({a}^{b}\right) = b \ln \left(a\right)$, this function can be rewritten as

$f \left(x\right) = \frac{1}{2} \ln \left(\arcsin \left({e}^{3 x}\right)\right)$

From here, the first issue is the natural logarithm. According to the chain rule, we know that $\frac{d}{\mathrm{dx}} \left[\ln \left(u\right)\right] = \frac{1}{u} \cdot u '$, and we have $u = \arcsin \left({e}^{3 x}\right)$.

$f ' \left(x\right) = \frac{1}{2} \left(\frac{1}{\arcsin} \left({e}^{3 x}\right)\right) \frac{d}{\mathrm{dx}} \left[\arcsin \left({e}^{3 x}\right)\right]$

Now, to differentiate the $\arcsin$ function, we will again use the chain rule, in that $\frac{d}{\mathrm{dx}} \left[\arcsin \left(u\right)\right] = \frac{1}{\sqrt{1 - {x}^{2}}} \cdot u '$, and this time $u = {e}^{3 x}$.

$f ' \left(x\right) = \frac{1}{2 \arcsin \left({e}^{3 x}\right)} \left(\frac{1}{\sqrt{1 - {\left({e}^{3 x}\right)}^{2}}}\right) \frac{d}{\mathrm{dx}} \left[{e}^{3 x}\right]$

Before moving on to find the derivative of ${e}^{3 x}$, simplify a little...

$f ' \left(x\right) = \frac{1}{2 \sqrt{1 - {e}^{6}} \arcsin \left({e}^{3 x}\right)} \cdot \frac{d}{\mathrm{dx}} \left[{e}^{3 x}\right]$

This final bout of differentiation also requires the chain rule, but differentiating things to the $e$ power is e-asy: $\frac{d}{\mathrm{dx}} \left[{e}^{u}\right] = {e}^{u} \cdot u '$, where $u = 3 x$.

$f ' \left(x\right) = \frac{1}{2 \sqrt{1 - {e}^{6}} \arcsin \left({e}^{3 x}\right)} \left({e}^{3 x} \frac{d}{\mathrm{dx}} \left[3 x\right]\right)$

Since $\frac{d}{\mathrm{dx}} \left[3 x\right] = 3$, this all simplifies to be

$f ' \left(x\right) = \frac{3 {e}^{3 x}}{2 \sqrt{1 - {e}^{6}} \arcsin \left({e}^{3 x}\right)}$