Hello Friends now we see the theorem which
is a base on derivative of quotient if
U and V are differentiable function of
X and if y equal to U upon V then DY
upon DX is equal to V into D u by DX
minus U into DV by DX whole thing
divided by v square provided V naught
equal to 0 now let's see the proof first
of all let Delta u Delta V Delta Y and
Delta X be the small increments in the
values of U V Y and X respectively
since Y is equal to u upon V therefore Y
change to y plus Delta Y is equal to u
change to u plus Delta U and V changes
to V Plus Delta V therefore Delta Y is
equal to u plus Delta u upon V Plus
Delta V minus y therefore Delta Y is
equal to tu plus Delta u upon V Plus
Delta V minus four by we write u upon V
since Y is equal to tu upon V the next
step will be Delta Y is equal to now
here we go for cross multiply so on
cross multiplying we get V in bracket u
plus Delta u minus u in bracket V Plus
Delta V upon V into V Plus Delta V so
the next step will be Delta Y is equal
to now on opening the bracket we have
been into u that is UV plus V Delta u
then minus UV and minus u Delta V
the whole thing / be in bracket V Plus
Delta V now here in the operator we
cancel Plus u B minus u v so the next
step will be Delta Y is equal to V into
Delta u minus u into Delta V upon V
square plus V into Delta V now dividing
both sides by Delta X I am taking limit
as Delta X tends to 0 we get
debate delta x tends to 0 Delta y upon
Delta X is equal to B into limit Delta X
tends to 0 Delta u upon Delta X minus u
into limit Delta X tends to 0 Delta V
upon Delta X and the whole thing divided
by B square plus v into limit Delta X
tends to 0 Delta V upon Delta X into
Delta X this we denote as a equation 1
now since U and V are differentiable
functions of X
therefore limit delta x tends to 0 Delta
u upon Delta X is equal to D u by DX and
limit Delta H tends to 0 Delta V upon
Delta X is equal to DV upon DX it means
the limit on right-hand side of equation
one exists therefore limit on left hand
side of equation one also exists and is
given by limit Delta H tends to 0 Delta
Y upon Delta X is equal to dy upon DX so
combining these all results in equation
one we get dy upon DX is equal to V into
D u by DX minus u into DV by DX upon v
square plus V
into DV by DX into 4 Delta X we write
zero therefore dy upon DX is equal to V
into D u by DX minus u into DV by DX
upon v square plus now V into D u by DX
into 0 so that u 0 provided V naught
equal to 0 so in this way finally we get
dy upon DX is equal to V into D u by DX
minus u into DV by DX upon v square
comma V naught equal to 0 so this is a
required proof for the given theorem
thank you
