Question

How do you differentiate `(x-cosx)/ (sinx+x)` using the quotient rule?

Answer 1

`dy/dx={1+x(sinx-cosx)+sinx+cosx}/(sinx+x)^2`

Explanation:

Let `y=(x-cosx)/(sinx+x)` Using Quotient Rule for Diffn., we have,
`dy/dx={(sinx+x)(x-cosx)'-(x-cosx)(sinx+x)'}/(sinx+x)^2`
`={(sinx+x)(1+sinx)-(x-cosx)(cosx+1)}/(sinx+x)^2`
`={sin^2x+xsinx+sinx+x+(cosx-x)(cosx+1)}/(sinx+x)^2`
`=(sin^2x+xsinx+sinx+x+cos^2x-xcosx+cosx-x)/(sinx+x)^2`
`={1+x(sinx-cosx)+sinx+cosx}/(sinx+x)^2`