Question

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

Answer 1

`2*y*dy/dx = 1/cos^2(x)`

Explanation:

y is an (unknown) function of x

Derive the left hand side
`d/dx(1) = 0`

Derive the right hand side

`d/dx(tan(x) - y^2) =d/dx(tan(x) - d/dx(y^2)`
Sum rule in differention

`tan(x) = sin(x)/cos(x)`
`d/dx(tan(x)) = d/dx(sin(x)/cos(x))`
`d/dx(tan(x)) =(sin(x)*sin(x) - (-cos(x))*cos(x))/cos(x)^2`
`d/dx(tan(x)) =(sin^2(x) +cos^2(x))/cos^2(x)`
`sin^2(x) +cos^2(x) = 1`
`d/dx(tan(x)) = 1/cos^2(x)`

`d/dx(y^2) = 2*y*d/dx(y) = 2*y*dy/dx`
The chain rule.

d/dx(tan(x)) = 1/cos^2(x) - 2ydy/dx #

Left hand side = right hand side

`0 = 1/cos^2(x) - 2*y*dy/dx`

Rewrite

`2*y*dy/dx = 1/cos^2(x)`