lpx88 SG I know how to use product rule..but final ans just cant get..ans is -2x (xsin2x - cos2x) Think is the factorisation part...
unclebutcher SG 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!
unclebutcher SG 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.
lpx88 SG 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
unclebutcher SG gd luck for prelims....btw is my answer correct? i make alot of careless mistakes too...
unclebutcher SG 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...
nightzip SG Yes, to add in some more steps, 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)
lpx88 SG HOLY CRAP U not careless Im the careless one-.- Thx man..Think my mind is just sleep deprived..
unclebutcher SG lol i finished the h2 maths differentiation topic abt 2 wks ago so still fresh i guess….
^tamago^ SG d/dx[x²cos²x] = x²[½(cos2x+1)] (by trigonometric identity) = ½x² + ½x²cos2x = x + xcos2x + (½x²)(-2sin2x) (by product rule) = x + xcos2x - x²sin2x = x(1+cos2x+xsin2x) d/dx[x²cos²x] = 2xcos²x + x²[d/dx(cos²x)] (by product rule) = 2xcos²x + x²[2(-sinx)(cosx)] = 2xcos²x - 2x²sinxcosx = 2x(cosx)(cosx-xsinx)
ispyyy SG 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....
