Question

How do you differentiate `y=cos^-1(1-3x^2)`?

Answer 1

`dy/dx=(6x)/sqrt(6x-9x^2)`

Explanation:

Easiest method is to rewrite;

`y=cos^-1(1-3x^2)` as `cosy=1-3x^2`

and to use implicit differentiation to give:
`-sinydy/dx =-6x`
`:. sinydt/dx =6x`
`:. dy/dx =(6x)/siny`

And using `cos^2A+sin^2A-=1 => siny =sqrt(1-cos^2y)`
`:. siny =sqrt(1-(1-3x^2)^2)`
`:. siny =sqrt(1-(1-3x^2)^2)`
`:. siny =sqrt(1-(1-6x+9x^2))`
`:. siny =sqrt(1-1+6x-9x^2)`
`:. siny =sqrt(6x-9x^2)`

And so, `dy/dx=(6x)/sqrt(6x-9x^2)`