Differentiation
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With respect to X,differentiate:
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erm..product rule?
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product rule.
den for cos^2 X, use chain rule.
verify this for me~.... -
I know how to use product rule..but final ans just cant get..ans is
-2x (xsin2x - cos2x)
Think is the factorisation part...
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x^2 d/dx (cos x)^2 + (cos x)^2 d/dx x^2
=
x^2 (2cosx(-sinx)) +(cosx)^2(2x)
simplify(i too lazy to do)
and:
voila! -
erm its prelims next month for you and you still can't get this right.....Originally posted by lpx88:I know how to use product rule..but final ans just cant get..ans is
-2x (xsin2x - cos2x)
Think is the factorisation part...
better study harder and spend less time on sgf!!!jkjk. -
No the thing is i alrdy know how to differentiate..but somehow i factorise here and there still cant get ans leh..so come here lorOriginally posted by unclebutcher:erm its prelims next month for you and you still can't get this right.....
better study harder and spend less time on sgf!!!jkjk. -
gd luck for prelims....btw is my answer correct? i make alot of careless mistakes too...Originally posted by lpx88:No the thing is i alrdy know how to differentiate..but somehow i factorise here and there still cant get ans leh..so come here lor -
Um..can u show me how u factorise till dat final ans?Originally posted by unclebutcher:gd luck for prelims....btw is my answer correct? i make alot of careless mistakes too...
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x^2.cos^2 x=
x^2(cosx)^2
d/dx x^2(Cosx)^2=
x^2 d/dx (cosx)^2 + (cosx)^2 d/dx x^2=
x^2 (2)(1)(cosx)(-sinx) + (cosx)^2(2x)=
-2x^2(cosxsinx) + 2x(cosx)^2=
2xcosx(cosx - xsinx)(answer)
see if got careless mistake... -
Yes, to add in some more steps,Originally posted by unclebutcher:x^2 d/dx (cos x)^2 + (cos x)^2 d/dx x^2
=
x^2 (2cosx(-sinx)) +(cosx)^2(2x)
simplify(i too lazy to do)
and:
voila!
x^2(-2cosxsinx) + (cos 2x+1)/2 (2x) <=== this is from cos2x= 2(cos^2 x) -1
=-2(x^2)(sin2x) + x(cos2x+1)
= -2x (xsin2x - cos 2x - 1) -
HOLY CRAPOriginally posted by unclebutcher:x^2.cos^2 x=
x^2(cosx)^2
d/dx x^2(Cosx)^2=
x^2 d/dx (cosx)^2 + (cosx)^2 d/dx x^2=
x^2 (2)(1)(cosx)(-sinx) + (cosx)^2(2x)=
-2x^2(cosxsinx) + 2x(cosx)^2=
2xcosx(cosx - xsinx)(answer) see if got careless mistake...
U not careless
Im the careless one-.-
Thx man..Think my mind is just sleep deprived.. -
lol i finished the h2 maths differentiation topic abt 2 wks ago so still fresh i guess....
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Thx evry1...i go sleep now...Cya all 2moro
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d/dx[x²•cos²x]
= x²[½(cos2x+1)] (by trigonometric identity)
= ½x² + ½x²•cos2x
= x + x•cos2x + (½x²)(-2sin2x) (by product rule)
= x + x•cos2x - x²sin2x
= x(1+cos2x+x•sin2x)
d/dx[x²•cos²x]
= 2x•cos²x + x²[d/dx(cos²x)] (by product rule)
= 2x•cos²x + x²[2(-sinx)(cosx)]
= 2x•cos²x - 2x²•sinx•cosx
= 2x(cosx)(cosx-x•sinx) -
let y be xcosx
let p be y^2
dy/dx = cosx - xsinx
dp/dy = 2y
Using chain rule,
ans = (cosx-xsinx)2y
muahaahaaaa.... my 1 rule.... -
also can...but why bother on such an easy question?Originally posted by ispyyy:let y be xcosx
let p be y^2
dy/dx = cosx - xsinx
dp/dy = 2y
Using chain rule,
ans = (cosx-xsinx)2y
muahaahaaaa.... my 1 rule.... -
lol >< LolOriginally posted by unclebutcher:also can...but why bother on such an easy question? -
yeah, brute force faster.Originally posted by unclebutcher:also can...but why bother on such an easy question?
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i'm sure.Originally posted by ^tamago^:yeah, brute force faster.