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

Jan 20, 2016

$f ' \left(x\right) = \frac{- x {e}^{2}}{\arcsin \left({e}^{2 - {x}^{2}}\right) \sqrt{{e}^{2 {x}^{2}} - {e}^{4}}}$

Explanation:

First, recognize that this can be written as

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

since the square root is really a power of $1 \text{/} 2$, and powers can be brought out of logarithms.

The first issue is the natural logarithm. But, according to the chain rule, $\frac{d}{\mathrm{dx}} \left[\ln \left(u\right)\right] = \frac{u '}{u}$. Thus,

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

The next issue lies in differentiating the arcsine function. The following rule can be used: $\frac{d}{\mathrm{dx}} \left[\arcsin \left(u\right)\right] = \frac{u '}{\sqrt{1 - {u}^{2}}}$

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

To differentiate the $e$ power, use the chain rule once more: $\frac{d}{\mathrm{dx}} \left[{e}^{u}\right] = {e}^{u} \cdot u '$

$f ' \left(x\right) = \frac{1}{2 \arcsin \left({e}^{2 - {x}^{2}}\right)} \cdot \frac{\frac{d}{\mathrm{dx}} \left[2 - {x}^{2}\right] {e}^{2 - {x}^{2}}}{\sqrt{1 - {e}^{4 - 2 {x}^{2}}}}$

$f ' \left(x\right) = \frac{1}{2 \arcsin \left({e}^{2 - {x}^{2}}\right)} \cdot \frac{- 2 x {e}^{2 - {x}^{2}}}{\sqrt{1 - {e}^{4 - 2 {x}^{2}}}}$

Nasty simplification from here on out, now that the calculus is over...

f'(x)=(-xe^(2-x^2))/(arcsin(e^(2-x^2))sqrt(1-e^(4-2x^2))

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

$f ' \left(x\right) = \frac{\frac{- x {e}^{2}}{{e}^{{x}^{2}}}}{\arcsin \left({e}^{2 - {x}^{2}}\right) \sqrt{\frac{{e}^{2 {x}^{2}} - {e}^{4}}{e} ^ \left(2 {x}^{2}\right)}}$

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

$f ' \left(x\right) = \frac{- x {e}^{2}}{\arcsin \left({e}^{2 - {x}^{2}}\right) \sqrt{{e}^{2 {x}^{2}} - {e}^{4}}}$