How do you differentiate this?

Question

#y=[ln(x+1)]^-2#

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Answer 1 · 2018-03-22T16:42:17

#(dy)/(dx)=-2/(x+1)[ln(x+1)]^(-3)#

We use chain rule here

As #y=[ln(x+1)]^(-2)#

#(dy)/(dx)=-2[ln(x+1)]^(-3)*1/(x+1)*1#

= #-2/(x+1)[ln(x+1)]^(-3)#