Derivative of fraction like equation = (vdu - udu) / v^2wherein,
u is numerator
v is denominator
kaya mo yan,
i-familiarize mo lang mga basic Formula on Derivative,
like:
d(uv) = u[d(v)] + v[d(u)]
d(sinu) = cosu [d(u)]
etc.
May libro ka naman ata dyan para maliwanagan ka hehe
maraming maraming salamat po, intindihin ko na lang po.y=sin^2x/cos^2x
y'=(cos^2x)(2sinx)(cosx)-(sin^2x)(2cosx)(-sinx)
y'= [cos^3x(2sinx)-(sin^3x)(2cosx)] / (cos^2x)^2
y=sin(2x)+cos^2x
y'=cos(2x)(2)+(2cosx)(-sinx)
y'=2cos2x - 2sinxcosx
dun po sa pangalawa pede po bang ang sagot ay 2cos2x-sin2xy=sin^2x/cos^2x
y'=(cos^2x)(2sinx)(cosx)-(sin^2x)(2cosx)(-sinx)
y'= [cos^3x(2sinx)-(sin^3x)(2cosx)] / (cos^2x)^2
y=sin(2x)+cos^2x
y'=cos(2x)(2)+(2cosx)(-sinx)
y'=2cos2x - 2sinxcosx
Hahaha, medyo mahirap pero masaya po pag natutunan niyo na.ay wala man lang akung na itindihan dito
ganon ba ...sana matotonan ko yanHahaha, medyo mahirap pero masaya po pag natutunan niyo na.
Oo bro same lang yan naka'double angle lang.dun po sa pangalawa pede po bang ang sagot ay 2cos2x-sin2x