Question

Find dy/dx of x cos y= sin (x+y) ?

Answer 1

`dy/dx=(xtanx+1-secx)secxcos^2y`.

Explanation:

We will use the Usual & the Rule of Implicit Differentiation.

`xcosy=sin(x+y)=sinxcosy+cosxsiny`,

`:. (x-sinx)cosy=cosxsiny`.

Dividing by `cosx*cosy`, we get,

`(x-sinx)/cosx=siny/cosy`.

`:. x/cosx-sinx/cosx=siny/cosy, i.e.,`

`tany=xsecx-tanx`.

`:. d/dx(tany)=d/dx(xsecx-tanx)`.

`:. sec^2y*dy/dx=x(secxtanx)+1*secx-sec^2x`,

`=(xtanx+1-secx)secx`.

`rArr dy/dx={(xtanx+1-secx)secx}/sec^2y, or,`

`dy/dx=(xtanx+1-secx)secxcos^2y`.

Enjoy Maths.!

Answer 2

This type of question must be of the form :"If `xcosy=sin(x+y)`,then prove that `(dy)/(dx)=(given)`.otherwise there are different answers.
`(dy)/(dx)=(cosx+xsinx-1)/(x^2-2xsinx+1)`

Explanation:

`color(blue)(d/(dx)(tan^(-1)X)=1/(1+X^2)`
`xcosy=sin(x+y)=sinxcosy+cosxsiny`
`=>xcosy-sinxcosy=cosxsiny`,
dividing both sides by `cosy!=0,` we get
`x-sinx=cosxtany=>tany=(x-sinx)/cosx`
`=>y=tan^(-1)((x-sinx)/cosx)`, Let, `color(red)(u=(x-sinx)/cosx)`, then
`(dy)/(dx)=1/(1+((x-sinx)/cosx)^2)(cosx(1-cosx)-(x-sinx)(-sinx))/(cos^2x)`
`=cancel((cos^2x))/(cos^2x+(x-sinx)^2)*(cosx-cos^2x+xsinx-sin^2x)/cancel((cos^2x))`
`=(cosx+xsinx-color(red)((cos^2x+sin^2x)))/(color(red)(cos^2x)+x^2-2xsinx+color(red)(sin^2x))`
`=(cosx+xsinx-1)/(x^2-2xsinx+1)`

Answer 3

Image reference...

Explanation:

Hope it helps....
Thank you...