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find dy/dx.. y= 3/8 (x) + 3/8 (sinxcosx) + 1/4 (cos^3 x sinx)

Mathematics
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\[\frac{ 3 }{ 8 }x + \frac{ 3 }{ 8 }sinxcosx + \frac{ 1 }{ 4 }\cos^3x sinx\]
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the final answer should be cos^4 x

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Other answers:

it is indeed cos^4 x still interested in finding how ?
yep.. can u help me @hartnn
sure, u know the product rule in differentiation ?
yes..
then tell me what u got the derivative of \(cos^3x sin x=?\)
\[(\cos^3x) (\cos x) + (\sin x)(3)(\cos^2 x)(-\sin x)\]
that is correct :) now u know the formula of sin 2x ?
\[\sin 2x= 2sinxcosx\]
am i right?
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 ?
where did u get 6/16 cos 2x??
\(\huge \frac{d}{dx}sin2x=2cos2x\)
ok.. and then.. what's next??
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 ??
there are 3 formulas for cos 2x
right, lets get everything in terms of cos....so which formula has cos in it ?
cos 2x= cos^2 x - sin^2 x cos 2x= 2cos^2 x - 1 these 2 formulas have cos
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
hmmm...can i cancell both (3/4)cos^2x ???
hmm ok i'll get it ... cos^4 x will be left.
but did u get how i got that step ? ask if there is any doubt in any step/term.
yes.. I get it.. but about this.. sin^2 x = 1- cos^2 x where din u get it?/
basic trignometric identity \(\huge \color{red}{sin^2x+cos^2x=1}\)
ow... Thank You For your help :))
welcome :)
^^

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