How do you differentiate #f(x)=tan(lnx) # using the chain rule?

1 Answer
Feb 16, 2016

Answer:

#f'(x)=sec^2(ln(x))/x#

Explanation:

The chain rule states that

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

First, we must know that the derivative of #tan(x)# is #sec^2(x)#. With this knowledge, we can create a version of the chain rule specific to tangent functions:

#d/dx(tan(x))=sec^2(x)#

#=>d/dx(tan(g(x))=sec^2(g(x))*g'(x)#

So, if we are differentiating the function #f(x)=tan(ln(x))#, we see that

#f'(x)=sec^2(ln(x))*d/dx(ln(x))#

We must now know that the derivative of #ln(x)# is #1"/"x#, yielding the simplified derivative:

#f'(x)=sec^2(ln(x))*1/x#

#f'(x)=sec^2(ln(x))/x#