Question

How do you implicitly differentiate `2=e^(xy)cosy`?

Answer 1

`y/(tan y - x)`

Explanation:

You could differentiate right away, but to make things simpler let us take the natural logarithm on both sides:

`ln 2 = xy + ln (cos y)`

Differentiating with respect to `x` on both sides:

`0 = y + x dy/dx + 1/(cos y) * (–sin y) * dy/dx`
`0 = y + x dy/dx - tan y * dy/dx`
`dy/dx = y/(tan y - x)`