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

1 Answer
Jan 20, 2016

Answer:

#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))#