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What are the first three derivatives of #(xcos(x)-sin(x))/(x^2)#?

Answer 1

Aurora Alvarado

The answer is:

#y''=(-x^3cosx+3x^4sinx+6xcosx-6sinx)/x^4#.

This is why:

#y'=(((cosx+x*(-sinx)-cosx)x^2-(xcosx-sinx)*2x))/x^4=#

#=(-x^3sinx-2x^2cosx+2xsinx)/x^4=#

#=(-x^2sinx-2xcosx+2sinx)/x^3#

#y''=((-2xsinx-x^2cosx-2cosx-2x(-sinx)+2cosx)x^3-(-x^2sinx-2xcosx+2sinx)*3x^2)/x^6=#

#=((-x^2cosx)x^3+3x^4sinx+6x^3cosx-6x^2sinx)/x^6=#

#=(-x^3cosx+3x^4sinx+6xcosx-6sinx)/x^4#.

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Answer 2

William Abdalla

The first three derivatives of the function ( \frac{xcos(x)-sin(x)}{x^2} ) are:

( f'(x) = \frac{-xsin(x)}{x^2} + \frac{cos(x)}{x} + \frac{2xcos(x) - 2sin(x)}{x^3} )
( f''(x) = \frac{-2cos(x)}{x} - \frac{2xcos(x) - 2sin(x)}{x^3} - \frac{3cos(x)}{x^2} + \frac{2xsin(x)}{x^3} - \frac{2cos(x)}{x^2} )
( f'''(x) = \frac{4sin(x)}{x} - \frac{2xsin(x)}{x^3} - \frac{6cos(x)}{x^2} - \frac{4xcos(x) - 4sin(x)}{x^3} + \frac{6cos(x)}{x^2} - \frac{6cos(x)}{x^3} )

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.