How do you differentiate #f(x)=ln(3(e^(sin^2x))^4)# using the chain rule?

1 Answer
Lotusbluete
Nov 9, 2015

I would start with something entirely different than the chain rule.
I would start with simplifying this expression as it is before differentiating.

Remember two rules:
#ln(a * b) = ln(a) + ln(b)#
#ln(a^b) = b * ln(a)#

Using this rules, the expression can be simplifyed in quite a drastical way:

#ln(3(e^(sin^2 x))^4) #
# = ln(3) + ln((e^(sin^2 x))^4)#
# = ln(3) + 4 * ln(e^(sin^2 x))#
...remember that #ln# and #e# are inverse functions...
# = ln(3) + 4 * sin^2(x)#

Now, this is something that can be differentiated in a reasonably easy way - better than the former one anyway.

#f(x) = ln(3) + 4 * sin^2(x) = ln(3) + 4 * u^2(x)#
where #u(x) = sin(x)#.

As you know, #u'(x) = cos(x)#, so using the chain rule we get:

#f'(x) = 0 + 4 * 2 u(x) * u'(x)#
#= 4 * 2 * sin(x) * cos(x)#
# = 8 sin(x) cos(x)#

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