Question

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

Answer 1

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

Explanation:

First, recognize that this can be written as

`f(x)=1/2ln(arcsin(e^(2-x^2)))`

since the square root is really a power of `1"/"2`, and powers can be brought out of logarithms.

The first issue is the natural logarithm. But, according to the chain rule, `d/dx[ln(u)]=(u')/u`. Thus,

`f'(x)=1/2(d/dx[arcsin(e^(2-x^2))])/arcsin(e^(2-x^2))`

The next issue lies in differentiating the arcsine function. The following rule can be used: `d/dx[arcsin(u)]=(u')/sqrt(1-u^2)`

`f'(x)=1/(2arcsin(e^(2-x^2)))*(d/dx[e^(2-x^2)])/sqrt(1-(e^(2-x^2))^2)`

To differentiate the `e` power, use the chain rule once more: `d/dx[e^u]=e^u*u'`

`f'(x)=1/(2arcsin(e^(2-x^2)))*(d/dx[2-x^2]e^(2-x^2))/sqrt(1-e^(4-2x^2))`

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

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

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

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