How do you differentiate $y=tan[ln(1-3x)]$ ?

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Answer 1 · 2017-12-29T21:58:25

$d/dx[y(u)]=(-3sec^2(ln(1-3x)))/(1-3x)$

Apply The Chain Rule:

Let $color(red)(u=ln(1-3x)$ And rewrite the problem

$y=tan[ln(1-3x)]->y=tan(color(red)u)$

By the chain rule: $d/dx[y(u)]=y'(u)*u'$

$y'=sec^2(color(red)u)$

$color(red)(u'=1/(1-3x)*-3=-3/(1-3x)$

$:.d/dx[y(u)]=sec^2(color(red)(ln(1-3x)))*color(red)(-3/(1-3x))=(-3sec^2(ln(1-3x)))/(1-3x)$

How do you differentiate y=tan[ln(1-3x)]y=tan[ln(1-3x)]?