Question

What is the implicit derivative of `1= e^y-xcos(xy)`?

Answer 1

`(dy)/dx=(cosxy-xysinxy)/(e^y+x^2(sinxy))`

Explanation:

`1=e^y−xcos(xy)`
`rArr(d1)/dx=d/dx(e^y−xcos(xy))`
`rArr0=(de^y)/dx-(d(xcos(xy)))/dx`
`rArr0=(dy/dx)e^y-(((dx)/dx)cosxy+x(dcosxy)/dx)`
`rArr0=(dy/dx)e^y-(cosxy+x(dxy)/dx(-sinxy))`
`rArr0=(dy/dx)e^y-(cosxy+x((y+x(dy)/dx)(-sinxy)))`
`rArr0=(dy/dx)e^y-(cosxy+x(-ysinxy-x(dy)/dx(sinxy)))`
`rArr0=(dy/dx)e^y-(cosxy-xysinxy-x^2(dy)/dx(sinxy))`
`rArr0=(dy/dx)e^y-cosxy+xysinxy+x^2(dy)/dx(sinxy)`
`rArr0=(dy/dx)e^y+x^2(dy)/dx(sinxy)-cosxy+xysinxy`
`rArr0=(dy/dx)(e^y+x^2(sinxy))-cosxy+xysinxy`
`rArrcosxy-xysinxy=(dy/dx)(e^y+x^2(sinxy))`

`rArr(dy)/dx=(cosxy-xysinxy)/(e^y+x^2(sinxy))`