Question

How do you implicitly differentiate # x cos(y) + y cos(x) = 1 ?

Answer 1

`dy/dx=frac(ysin(x)-cos(y))(cos(x)-xsin(y))`

Explanation:

I'm assuming we're differentiating with respect to `x`.

`d/dx[xcos(y)+ycos(x)=1]`

Lots of use of the Product Rule and Chain Rule:
`cos(y)-xsin(y)[dy/dx]+[dy/dx]cos(x)-ysin(x)=0`
`[dy/dx]cos(x)-xsin(y)[dy/dx]=ysin(x)-cos(y)`
`dy/dx(cos(x)-xsin(y))=ysin(x)-cos(y)`
`dy/dx=frac(ysin(x)-cos(y))(cos(x)-xsin(y))`