How do you use the chain rule to differentiate $ln (tan x)$ $ln(tanx)$ ?

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Answer 1 · 2017-01-08T04:43:19

$\frac{d y}{d x} = \frac{{sec}^{2} x}{tan x}$ $(dy)/(dx)=(sec^2x)/tanx$

We can treat $f (x) = ln (tan x)$ $f(x)=ln(tanx)$ as a function in terms of $tan x$ $tanx$ .

Let $u = tan x$ $u=tanx$ , $y = ln u$ $y=lnu$

Chain rule: $\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $(dy)/(dx)=(dy)/(du)*(du)/(dx)$

$\frac{d y}{d u} = \frac{1}{u}$ $(dy)/(du)=1/u$

$\frac{d u}{d x} = {sec}^{2} x$ $(du)/(dx)=sec^2x$

$:.(dy)/(dx)=1/(tanx)*sec^2x$

$=(sec^2x)/(tanx)$

Further simplifying...

$=(1/(cos^2x))/((sinx)/(cosx))$

$=1/(sinxcosx)$

$=2/(2sinxcosx)$

$=2/(sin2x)$

$=2csc2x$

How do you use the chain rule to differentiate ln(tanx)ln(tanx)ln(tanx)?