How do you differentiate $(sinx+sinxcosx)/x$ ?

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Answer 1 · 2017-11-22T17:37:52

$(x(cosx+cos^2x-sin^2x)-sinx-sinxcosx)/x^2$

$d/(dx)(u/v)=(vu'-uv')/v^2$

$d/(dx)(uv)=vu'+uv'$

$d/(dx)(((sinx+sinxcosx))/x)$

$=(xd/(dx)(sinx+sinxcosx)-(sinx+sinxcosx)d/(dx)(x))/x^2$

we will use the product rule on $sinxcosx$

$=(x(cosx+cosxcosx-sinxsinx)-(sinx+sinxcosx))/x^2$

$=(x(cosx+cos^2x-sin^2x)-sinx-sinxcosx)/x^2$

How do you differentiate (sinx+sinxcosx)/x(sinx+sinxcosx)/x?