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

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Answer 1 · 2016-05-07T14:09:49

#-1/2e^(2-x)/(arcsin(e^(2-x))sqrt(1-e^(4-2x)))#

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Draw a triangle to express your inverse trigonometric term.
We find that #e^(2-x)=siny# or #arc(e^(2-x))=y#.

We want to find #(df)/(dx)# which also equals to #(df)/(dy)(dy)/(dx)#.

#(df)/(dy)=d/(dy)ln(sqrty)=1/(2y)#

Differentiate explicitly:
#e^(2-x)=siny#
#-e^(2-x)dx=(cosy)dy# or #(dy)/(dx)=-e^(2-x)/cosy#

Therefore, your answer is;
#(df)/(dx)=(df)/(dy)(dy)/(dx)#.
#=1/(2y)xx-e^(2-x)/cosy#
#=-1/2e^(2-x)/(arcsin(e^(2-x))sqrt(1-e^(4-2x)))#

Note:
#y=arc(e^(2-x))#
#cosy= sqrt(1-e^(4-2x))# (refer to the triangle)

Answer 2 · 2016-05-07T14:51:09

Just adding to Alexander's answer, #x>=2#.. .

#e^(2-x)>=0#. Yet, as it is sine of an angle, it has to be #<=1#. So, the exponent #2-x<=0#. Thus, #x>=2#.

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