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bii17
Group Title
find dy/dx..
y= 3/8 (x) + 3/8 (sinxcosx) + 1/4 (cos^3 x sinx)
 one year ago
 one year ago
bii17 Group Title
find dy/dx.. y= 3/8 (x) + 3/8 (sinxcosx) + 1/4 (cos^3 x sinx)
 one year ago
 one year ago

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bii17 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ 3 }{ 8 }x + \frac{ 3 }{ 8 }sinxcosx + \frac{ 1 }{ 4 }\cos^3x sinx\]
 one year ago

bii17 Group TitleBest ResponseYou've already chosen the best response.1
the final answer should be cos^4 x
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
it is indeed cos^4 x still interested in finding how ?
 one year ago

bii17 Group TitleBest ResponseYou've already chosen the best response.1
yep.. can u help me @hartnn
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
sure, u know the product rule in differentiation ?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
then tell me what u got the derivative of \(cos^3x sin x=?\)
 one year ago

bii17 Group TitleBest ResponseYou've already chosen the best response.1
\[(\cos^3x) (\cos x) + (\sin x)(3)(\cos^2 x)(\sin x)\]
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
that is correct :) now u know the formula of sin 2x ?
 one year ago

bii17 Group TitleBest ResponseYou've already chosen the best response.1
\[\sin 2x= 2sinxcosx\]
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
yes! u see how we can use that to differentiate middle term easily... 3/8 sin x cos x = 3/16 sin2x after diff. 6/16 cos2x = 3/8 cos 2x ok with this ?
 one year ago

bii17 Group TitleBest ResponseYou've already chosen the best response.1
where did u get 6/16 cos 2x??
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
\(\huge \frac{d}{dx}sin2x=2cos2x\)
 one year ago

bii17 Group TitleBest ResponseYou've already chosen the best response.1
ok.. and then.. what's next??
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
the first term diff. is just 3/8. now the 'differentiation' part is over.Lets put it all together and see what we can simplify: \(\large (3/8)+3/8cos2x(3/4)cos^2xsin^2x+(1/4)cos^4x\) now do u know the formula for cos2x ??
 one year ago

bii17 Group TitleBest ResponseYou've already chosen the best response.1
there are 3 formulas for cos 2x
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
right, lets get everything in terms of cos....so which formula has cos in it ?
 one year ago

bii17 Group TitleBest ResponseYou've already chosen the best response.1
cos 2x= cos^2 x  sin^2 x cos 2x= 2cos^2 x  1 these 2 formulas have cos
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
lets use 2nd one, where we directly get \(1+cos2x=2cos^2x\) and let use the fact that \(sin^2x=1−cos^2x\) so now we have : \((3/4)cos^2x−(3/4)cos^2x(1−cos^2x)+(1/4)cos^4x\) can u simplify this and tell me what gets cancelled
 one year ago

bii17 Group TitleBest ResponseYou've already chosen the best response.1
hmmm...can i cancell both (3/4)cos^2x ???
 one year ago

bii17 Group TitleBest ResponseYou've already chosen the best response.1
hmm ok i'll get it ... cos^4 x will be left.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
but did u get how i got that step ? ask if there is any doubt in any step/term.
 one year ago

bii17 Group TitleBest ResponseYou've already chosen the best response.1
yes.. I get it.. but about this.. sin^2 x = 1 cos^2 x where din u get it?/
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
basic trignometric identity \(\huge \color{red}{sin^2x+cos^2x=1}\)
 one year ago

bii17 Group TitleBest ResponseYou've already chosen the best response.1
ow... Thank You For your help :))
 one year ago
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