Question

If `3-x^2=tany`, then what is `dy/dx`?

Answer 1

The answer is `=(-2x)/(x^4-6x^2+10)`

Explanation:

We use

`(tanx)'=sec^2x`

`tan^2x+1=sec^2x`

`3-x^2=tany`

`sec^2y=1+tan^2y=1+(3-x^2)^2=1+9-6x^2+x^4`

Differentiating both sides

`-2x=sec^2y dy/dx`

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

`=(-2x)/(x^4-6x^2+10)`