Asked by Anonymous
Find the differential of y=ln(sqrt(4+t^2).
dy=?
dy=?
Answers
Answered by
Steve
since ln(sqrt(u)) = 1/2 ln u
y = 1/2 ln(4+t^2)
dy = 1/2 * 1/(4+t^2) * 2t dt
= t/(4+t^2) dt
y = 1/2 ln(4+t^2)
dy = 1/2 * 1/(4+t^2) * 2t dt
= t/(4+t^2) dt
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