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

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Answer 1 · 2017-12-29T21:58:25

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

Apply The Chain Rule:

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

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

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)]y=tan[ln(1-3x)]?