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

Question

40119 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-29T15:15:19

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:

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

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