How do you find the derivative of #ln(tanx)#?

1 Answer
Jim H
Apr 29, 2015

Use the chain rule and use #d/dx(lnu) = 1/u (du)/dx#.

We'll also need #d/dx(tanx) = sec^2x#

#d/dx(ln(tanx))=1/tanx d/dx(tanx) = 1/tanx sec^2x#

We are finished with the calculus, but we can rewrite the answer using trigonometry and algebra:

#d/dx(ln(tanx))= 1/(sinx/cosx) 1/(cos^2x)= 1/sinx 1/cosx = cscx secx#

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