# How do you differentiate f(x) = ln(1/sqrt(arcsin(e^(x)) ) )  using the chain rule?

##### 1 Answer
Aug 7, 2017

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

#### Explanation:

Using the chain rule we can differentiate a function of a function, so for example:

$\frac{d}{\mathrm{dx}} \left({\left(1 + 2\right)}^{10}\right) = 10 {\left(1 + 2 x\right)}^{9} \frac{d}{\mathrm{dx}} \left(1 + 2 x\right)$

Where we have implicitly used the chain rule:

$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) \setminus g ' \left(x\right)$

To give:

$\frac{d}{\mathrm{dx}} {X}^{n} = n {X}^{n - 1} \frac{\mathrm{dX}}{\mathrm{dx}}$ and $X = X \left(x\right)$

In this question the function has been specifically designed to test knowledge of the chain rule as it will need to be applied several times. We than join (or "chain") the results together to get the final derivative:

Note that we will need the following standard derivatives:

$\left[A\right] \setminus \frac{d}{\mathrm{dx}} \ln x \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{x}$
$\left[B\right] \setminus \frac{d}{\mathrm{dx}} \arcsin x = \frac{1}{\sqrt{1 - {x}^{2}}}$
$\left[C\right] \setminus \frac{d}{\mathrm{dx}} {e}^{x} \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {e}^{x}$

So we can simplify the expression using the log rules:

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

Chain rule application no 1 (using [A]) gives

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

Chain rule application no 2 (using [B]) gives

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

Chain rule application no 3 (using [C]) gives

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

Leading to the final result:

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