Differentiate [log squared (e^x.....)]
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Question:  differentiate [log^2 (e^x + x - 1)] .

Let u = ln(v), wher v = e^x + x - 1. Then

y = u^2

dy/dx = (dy/du)*(du/dv)*(dv/dx)

and

dy/du = 2u,

du/dv = 1/v,

dv/dx = e^x + 1.

Then,

dy/dx = (2u)*(1/v)*(e^x + 1) = (2ln(v))*(1/(e^x + x - 1))*(e^x + 1)

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

by Level 11 User (80.8k points)

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