# How do you differentiate f(x) = ln(sqrt(arcsin(e^(2-x)) )  using the chain rule?

May 7, 2016

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

#### Explanation:

Draw a triangle to express your inverse trigonometric term.
We find that ${e}^{2 - x} = \sin y$ or $a r c \left({e}^{2 - x}\right) = y$.

We want to find $\frac{\mathrm{df}}{\mathrm{dx}}$ which also equals to $\frac{\mathrm{df}}{\mathrm{dy}} \frac{\mathrm{dy}}{\mathrm{dx}}$.

$\frac{\mathrm{df}}{\mathrm{dy}} = \frac{d}{\mathrm{dy}} \ln \left(\sqrt{y}\right) = \frac{1}{2 y}$

Differentiate explicitly:
${e}^{2 - x} = \sin y$
$- {e}^{2 - x} \mathrm{dx} = \left(\cos y\right) \mathrm{dy}$ or $\frac{\mathrm{dy}}{\mathrm{dx}} = - {e}^{2 - x} / \cos y$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dy}} \frac{\mathrm{dy}}{\mathrm{dx}}$.
$= \frac{1}{2 y} \times - {e}^{2 - x} / \cos y$
$= - \frac{1}{2} {e}^{2 - x} / \left(\arcsin \left({e}^{2 - x}\right) \sqrt{1 - {e}^{4 - 2 x}}\right)$

Note:
$y = a r c \left({e}^{2 - x}\right)$
$\cos y = \sqrt{1 - {e}^{4 - 2 x}}$ (refer to the triangle)

May 7, 2016

Just adding to Alexander's answer, $x \ge 2$.. .

#### Explanation:

${e}^{2 - x} \ge 0$. Yet, as it is sine of an angle, it has to be $\le 1$. So, the exponent $2 - x \le 0$. Thus, $x \ge 2$.

Also, for differentiation, arc sine should be taken as a single-valued function..