Question

What is the implicit derivative of `1=ytanx-y^2`?

Answer 1

`dy/dx=(ysec^2x)/(2y-tanx)`

Explanation:

`d/dx[1=ytanx-y^2]`

Remember that taking the derivative of any term with a `y` when implicitly differentiating will put the chain rule into effect and split out a `dy/dx` term. Also remember that finding `d/dx[ytanx]` will require the product rule.

`0=tanxdy/dx+yd/dx[tanx]-2ydy/dx`

`0=tanxdy/dx+ysec^2x-2ydy/dx`

`2ydy/dx-tanxdy/dx=ysec^2x`

`dy/dx(2y-tanx)=ysec^2x`

`dy/dx=(ysec^2x)/(2y-tanx)`