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Differential Calculus

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5 Jan 2012 20:52:32 IST

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389

differentiate:

Mathematics

y=sinⁿxcosnx what is dy/dx?

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gsanjaychowdary

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5 Jan 2012 21:33:03 IST

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sinnx[-sinnx][n]+cosnx[cosnx][n]

Kriti Sinha

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5 Jan 2012 21:45:35 IST

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Starting off with, in sinnx, it is sin(^n)x [n is a power value)that is not the answer..the options are:a) nsin(^n-1)xcos(n+1)xb) nsin(^n-1)xsin(n+1)xc)nsin(^n-1)xcos(n-1)xd)nsin(^n-1)xcosnx

Kriti Sinha

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5 Jan 2012 21:48:45 IST

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Read this: the above has several typos;Starting off with, in sinnx, it is sin(^n)x [n is a power value]the options are: (a)nsin(^n-1)xcos(n+1)x (b)nsin(^n-1)xsin(n+1)x (c)nsin(^n-1)xcos(n-1)x (d)nsin(^n-1)xcosnx

Akshay I Parmar

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7 Jan 2012 16:41:37 IST

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dy/dx=1/2[n cos(nx+x) + cos(nx+x) -n cos(nx-x) + cos(nx-x)]

Aditya Sindhu Pandey

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13 Jan 2012 23:56:14 IST

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-nsinxsinnx+cosxcosnx

Bhupendra Ramesh Sharma

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28 Mar 2012 23:55:46 IST

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y = sinx.cosx

by using product rule

dy/dx = -sin²x + cos²x

dy/dx = cos²x - sin²x

Tarun Garg

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28 Mar 2012 23:59:31 IST

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cos2nx

Bhupendra Ramesh Sharma

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28 Mar 2012 23:59:38 IST

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y = sinx.cosnx

by using product rule

dy/dx = -nsinxsin(nx) + cosnx.cosx

Bhupendra Ramesh Sharma

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29 Mar 2012 00:00:28 IST

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y = sinx.cosnx

by using product rule

dy/dx = -nsinxsin(nx) + cosnx.cosx

Tarun Garg

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29 Mar 2012 00:08:35 IST

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sinⁿxcosⁿx=y

differntiate.

dy/dx=cosⁿx.(nsin^n-1xcosx)+sinⁿx(ncos^n-1x(-sinx))=nsin^n-1xcos^n-1x[cos²x.-sin²x]=nsin^n-1x.cos^n-1x.cos2x

Tarun Garg

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29 Mar 2012 00:13:41 IST

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i think question is let y=sinⁿx.cosnx

differentiate

dy/dx=cosnx[nsin^n-1xcosx]+sinⁿx[-nsinnx.]=nsin^n-1x{{cosnx.cosx-sinnx.sinx}}=nsin^n-1xcos(n+1)x..............................hhhhhuuuuuuuuuuuuurrrrrrrrrrrraaaaaaaaaaaaayyyyyyyyyyyyyyyyyyyyyyyyyyyyy ption a is correct so how many like for me

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