bii17
find dy/dx..
y= 3/8 (x) + 3/8 (sinxcosx) + 1/4 (cos^3 x sinx)
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bii17
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\[\frac{ 3 }{ 8 }x + \frac{ 3 }{ 8 }sinxcosx + \frac{ 1 }{ 4 }\cos^3x sinx\]
ali110
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bii17
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the final answer should be cos^4 x
hartnn
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it is indeed cos^4 x
still interested in finding how ?
bii17
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yep.. can u help me @hartnn
hartnn
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sure, u know the product rule in differentiation ?
bii17
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yes..
hartnn
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then tell me what u got the derivative of \(cos^3x sin x=?\)
bii17
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\[(\cos^3x) (\cos x) + (\sin x)(3)(\cos^2 x)(-\sin x)\]
hartnn
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that is correct :)
now u know the formula of sin 2x ?
bii17
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\[\sin 2x= 2sinxcosx\]
bii17
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am i right?
hartnn
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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 ?
bii17
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where did u get 6/16 cos 2x??
hartnn
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\(\huge \frac{d}{dx}sin2x=2cos2x\)
bii17
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ok.. and then.. what's next??
hartnn
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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 ??
bii17
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there are 3 formulas for cos 2x
hartnn
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right, lets get everything in terms of cos....so which formula has cos in it ?
bii17
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cos 2x= cos^2 x - sin^2 x
cos 2x= 2cos^2 x - 1
these 2 formulas have cos
hartnn
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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
bii17
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hmmm...can i cancell both (3/4)cos^2x ???
bii17
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hmm ok i'll get it ... cos^4 x will be left.
hartnn
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but did u get how i got that step ?
ask if there is any doubt in any step/term.
bii17
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yes.. I get it.. but about this.. sin^2 x = 1- cos^2 x where din u get it?/
hartnn
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basic trignometric identity
\(\huge \color{red}{sin^2x+cos^2x=1}\)
bii17
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ow... Thank You For your help :))
hartnn
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welcome :)
bii17
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^^