What is the derivative of $f(x)=tan(ln(cosx))$ ?

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Answer 1 · 2015-11-10T23:11:20

$dy/dx = -sec^2(ln(cos(x)))tan(x)$

So we have

$y = tan(ln(cos(x)))$

Let's say that $ln(cos(x)) = u$ , so we have

$y = tan(u)$

$dy/dx = d/(du)tan(u)(du)/dx$
$dy/dx = sec^2(u)d/dx(ln(cos(x))$

Let's say that $cos(x) = v$ , so we have

$dy/dx = sec^2(u)d/(dv)ln(v)(dv)/dx$
$dy/dx = sec^2(u)/vd/dxcos(x)$
$dy/dx = -(sec^2(u)sin(x))/v$

Putting everything in terms of $x$

$dy/dx = -sec^2(ln(cos(x)))sin(x)/cos(x)$
$dy/dx = -sec^2(ln(cos(x)))tan(x)$

What is the derivative of f(x)=tan(ln(cosx))f(x)=tan(ln(cosx))?