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Differential Calculus

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28 Mar 2012 23:38:25 IST

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376

Differentiate

Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Differential Calculus

y = sin2x.cos3x

Comments (5)

Tarun Garg

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28 Mar 2012 23:39:06 IST

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what to be do with this

Tarun Garg

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Joined: 21 Mar 2012

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28 Mar 2012 23:42:48 IST

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y=sin2x.cos3x=2sin2x.cos3x/2=sin(2x+3x)-sin(3x-2x)/2=sin5x-sinx/2

now

differentiate

dy/dx=co5x*5-cosx/2=5cos5x-cosx/2

Arjun Virmani

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29 Mar 2012 05:19:17 IST

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just diffferentiate using product rule..

y = f(x) . g(x)

y' = f' (x) . g(x) + f(x) . g'(x)

in this case f(x) = sin2x and g(x) = cos 3x

y' = 2cos2x cos 3x + sin2x (-3sin3x)

ans!!

rst rst

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29 Mar 2012 11:23:04 IST

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2cos3x.cos2x-3sin2xsin3x

rst rst

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Joined: 10 Sep 2011

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29 Mar 2012 11:26:26 IST

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its tooooo easy........only product rule n then func. of func inside it

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