i dunno how to solve this maths ... hiks .. im so desperate f(x) = ( 2x – 1 )² ( x + 2 ), f’(x) = …. f(x) = sin³ (3 – 2x). f’(x) = …. f(x) = sin² ( 2x + π/6 ), f′(0) = …. Plz help me .. deadline next week .. T____T
#1 = 12x^2 + 8x - 7 I'll edit this post with the rest of the answers when I'm done... It could be a few min its been a while since I've done differentiation EDIT: nvm SoC is going at it.. I need to do some reviewing >< I should be able to do this crap still ><
( 2x – 1 )² = 4x^2-4x+1 = 8x - 4 ( x + 2 ) = X+2 = 1 1x(4x^2-4x+1) + (X+2)(8x-4) 4x^2-4x+1 + 8x^2+12x-8 12x^2+8x-7. Yeh #1 is right (Not done 2/3 yet)
Doulbe post to alert person above me I meant by not done 2/3 that i haven't studied that, so ive no idea sorry Good luck
is it difficult ?? SoC haven't studied that ?? hmm hmm ... really ?? huhuhuhh ... its too difficult for me .. maths = 5*ck
Sin^3 and sin^2 isn't on our sylabus to learn. If i knew what sin^3 /Sin^2 differentated to, id be able to do it.
owh .. i see .. someone told me (whos living and school in NYC) .. the maths is more difficult in my country, Indonesia.
It all depends on the sylabus and exam board and your countries education policies. The UK was heavily messed up by margaret thatcher and a load of corrupt politicians who have delayed it and made the education system RUBBISH! What age are you doing this at?
same age then So its probably just different topics covered Roughly the same difficulty so it's not too hard then tbh
btw, thx for helping me (the 1st) .. and ill find out the answer for 2/3 someday, ill asking you for help ^^ .. but, how to translate it from my word to your word .. hehehehe ... thx SoC
yw and ive just realised the question is f(x) = sin³ (3 – 2x) meaning you can rewrite it as f(X) = sin(3-2x)^3 Which is then solveable, and you should be able to do via chain rule Hope that helps
I used to have a formula that worked for solving trig chains simply... gotta go back in time to find it ><