How do you differentiate #tan(3x^2) - csc ( ln(4x) )^2#?

1 Answer
Jun 23, 2015

Answer:

#6xsec^2(3x^2)+(2*csc^2(ln(4x))*cot(ln(4x)))/x#

Explanation:

Wow, this looks tough. Let's break it down piece by piece.

First, let's take a look at the #tan(3x^2)# term. We apply the chain rule when we differentiate this term.

We first differentiate this term with respect to the inner function #3x^2# and then multiply the result by the derivative of #3x^2# with respect to #x# (By the chain rule).

#(tan(3x^2))^' = sec^2(3x^2)*(6x)=6xsec^2(3x^2)#

Not too bad! :)
(Remember: #d/dx(tanx)=sec^2(x)#)

Now for the other (very scary) term, #-csc^2(ln(4x))#.

Wait a minute, that's not too bad. There are just four functions nested in each other; the #4x# in the #ln()#, the #ln()# in the #csc()# and finally the #csc()# in the #()^2#.

This means that we will need to apply the chain rule three times.

First we see that,

#(-csc^2(ln(4x)))^'=[-2csc(ln(4x))]*[csc(ln(4x))]^'#

Now,

# [csc(ln(4x))]^' = [-csc(ln(4x))*cot(ln(4x))][ln(4x)]^'#

and

#[ln(4x)]^' = [1/(4x)]*[4x]^' = 4(1/(4x))= 1/x#

Putting everything together we have

#-csc^2(ln(4x)))^'=([-2csc(ln(4x))]*[-csc(ln(4x))*cot(ln(4x))])/x#
#=(2*csc^2(ln(4x))*cot(ln(4x)))/x#

So finally, adding on #6xsec^2(3x^2)#, we have the derivative of the given expression as

#6xsec^2(3x^2)+(2*csc^2(ln(4x))*cot(ln(4x)))/x#

Woohoo! :)