Find the differential of y=ln(sqrt(4+t^2). dy=?

Science & Mathematics by Anonymous 2018-05-11 11:08:58

Social Science

5 answers

Anonymous

y=ln[sqr(4+t^2)] => dy=tdt/(4+t^2)
Anonymous

dy = (1/sqrt(4+t^2) ) d/dt ( sqrt(4+t^2)) dy = (1/sqrt(4+t^2)) 1/(2sqrt(4+t^2)) d/dt (4+t^2) dy = (1/sqrt(4+t^2)) 1/(2sqrt(4+t^2)) (2t) dy = t /((4+t^2)) dy = t/(4+t^2)
Anonymous

y=ln(sqrt(4+t^2) = (1/2)ln(4+t^2) dy = (t/(4+t^2))dt
Anonymous
Anonymous

I guess you mean y = ln(sqrt(4+t^2)). y = (1/2)*ln(4 + t^2), so dy/dt = (1/2)*(2t)/(4+t^2) = t/(4+t^2), and dy = t dt/(4 + t^2).