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

1 Answer
Jan 23, 2016

Answer:

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

Explanation:

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)))#