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#