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

1 Answer
Jun 23, 2015

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! :)