Question

How do you differentiate `y cos x = 1 + sin (xy)`?

Answer 1

`dy/dx = (y [cos (xy) + sin x])/(cos x - x cos (xy))`

Explanation:

Always think of your chain, product and quotient rules when differentiating

`y cos x = 1 + sin (xy)`
Diff wrt `x`:
`dy/dx cos x - y sin x = cos (xy) [y + x dy/dx]` (apply your rules)
`dy/dx cos x - y sin x = y cos (xy) + x cos (xy) dy/dx` (simplify)
`dy/dx cos x - x cos (xy) dy/dx = y cos (xy) + y sin x` (collect `dy/dx`)
`dy/dx [cos x - x cos (xy)] = y [cos (xy) + sin x]` (factorise)
`dy/dx = (y [cos (xy) + sin x])/(cos x - x cos (xy))` VOILA