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# find the diffrentiation of first principal esinx.cosx.

Saurabh Koranglekar
2 years ago
2075 Points
2 years ago
saurabh’s answer is NOT by first principle.
to use the first principle, we consider the ratio:
[esin(x+h).cos(x+h) – esinx.cosx]/h
=[esin(x+h).{cos(x+h) – cosx}+{esin(x+h) – esinx}cosx]/h
=[esin(x+h).{cos(x+h) – cosx}]/h+[{esin(x+h) – esinx}cosx]/h
now we set h tends to zero
=esin(x+0) Lt{cos(x+h) – cosx}/h + cosx Lt{esin(x+h) – esinx}/h
=esin(x+0) Lt2sin(x+h/2)sin(-h/2)/h + cosx*Lt{esin(x+h) – esinx}/h
=esinx( –sinx)+cosxLt{esin(x+h) – esinx}/h [here we have used the fact that lim x tends to zero sinx/x=1]
now let Lt{esin(x+h) – esinx}/h = m
then m=esinx Lt{esin(x+h) – sinx – 1}/{sin(x+h) – sinx}*Lt{sin(x+h) – sinx}/h
=esinx .1.cosx [here we have put sin(x+h) – sinx=y and then used the fact that lim x tends to zero (e^x-1)/x=1 and that differntiation of sinx is cosx which you can derive by using sina – sinb formula]
esinx( –sinx) + cosx*esinx .1.cosx
esinx (cos^2x – sinx)