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)#