Differentiation of sin[sin (sinx)]....plz give the right answer
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Plz mark as brainliest
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Hey!!!...Here is ur answer
d/dx [sin{sin (sinx)]
=> cos {sin (sinx)} d/dx [sin (sinx)]
=> cos (sinx).cos {sin (sinx)} d/dx (sinx)
=> cosx.cos (sinx).cos {sin (sinx)}
Hope it will help
d/dx [sin{sin (sinx)]
=> cos {sin (sinx)} d/dx [sin (sinx)]
=> cos (sinx).cos {sin (sinx)} d/dx (sinx)
=> cosx.cos (sinx).cos {sin (sinx)}
Hope it will help
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