How do you differentiate #f(x) = ln(sqrt(arcsin(e^(3x)) ) # using the chain rule?

The first step we should take is to simplify the function by using logarithm rules to remove the square root from the function.

Since the square root is really a #1"/"2# power, and #ln(a^b)=bln(a)#, this function can be rewritten as

#f(x)=1/2ln(arcsin(e^(3x)))#

From here, the first issue is the natural logarithm. According to the chain rule, we know that #d/dx[ln(u)]=1/u*u'#, and we have #u=arcsin(e^(3x))#.

#f'(x)=1/2(1/arcsin(e^(3x)))d/dx[arcsin(e^(3x))]#

Now, to differentiate the #arcsin# function, we will again use the chain rule, in that #d/dx[arcsin(u)]=1/(sqrt(1-x^2))*u'#, and this time #u=e^(3x)#.

#f'(x)=1/(2arcsin(e^(3x)))(1/sqrt(1-(e^(3x))^2))d/dx[e^(3x)]#

Before moving on to find the derivative of #e^(3x)#, simplify a little...

#f'(x)=1/(2sqrt(1-e^6)arcsin(e^(3x)))*d/dx[e^(3x)]#

This final bout of differentiation also requires the chain rule, but differentiating things to the #e# power is e-asy: #d/dx[e^u]=e^u*u'#, where #u=3x#.

#f'(x)=1/(2sqrt(1-e^6)arcsin(e^(3x)))(e^(3x)d/dx[3x])#

Since #d/dx[3x]=3#, this all simplifies to be

#f'(x)=(3e^(3x))/(2sqrt(1-e^6)arcsin(e^(3x)))#