How do you differentiate #f(x)=ln(tanx)# using the chain rule?

1 Answer
Dec 8, 2015

Answer:

#1/(tan(x)*cos^2(x))=csc(x)sec(x)#

Explanation:

You need the chain rule for composite functions, and the rule states that

#d/dx f(g(x)) = f'g(x) * g'(x)#

So, you differentiate the outer function, leaving its argument untouched, and then you multiply by the derivative of the inner function.

In your case, the outer function is #f(x)=ln(x)#, and the inner function is #g(x)=tan(x)#

So, the derivative of the outer function is #d/dx ln(x)=1/x#, so

#f'(g(x))=1/tan(x)#.

On the other hand, the derivative of #tan(x)# is #1/cos^2(x)#.

So,

#f'g(x) * g'(x) = 1/tan(x) * 1/cos^2(x) = 1/(tan(x)*cos^2(x))#

And since #tan(x)=sin(x)/cos(x)#, we have that

#tan(x)*cos^2(x)=sin(x)cos(x)#, and so the derivative becomes

#1/(sin(x)cos(x))=csc(x)sec(x)#