What is the derivative of #sinx / (x^2+sinx)#?

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Answer 1 · 2018-03-28T17:55:06

#(dy)/(dx)=(x^2cosx-2xsinx)/((x^2+sinx)^2)#

Let,

#y=sinx/(x^2+sinx#

We know that,

#color(red)(d/(dx)(u/v)=(v(du)/(dx)-u(dv)/(dx))/v^2#

So,

#(dy)/(dx)=((x^2+sinx)(cosx)-sinx(2x+cosx))/((x^2+sinx)^2)#

#=(x^2cosx+cancel(sinxcosx)-2xsinx- cancel(sinxcosx))/((x^2+sinx)^2)#

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