Question

How do you differentiate `y=csc^-1(4x^2)`?

Answer 1

`dy/dx=(-2)/(xsqrt(16x^4 - 1)) (=(-2)/(xsqrt((4x^2+1)(4x^2-1))))`

Explanation:

`y=csc^-1(4x^2) <=> cscy=4x^2`

Differentiating implicitly we have:
`cscy=4x^2`
`-cscycotydy/dx=8x`
`-(4x^2)cotydy/dx=8x`
`cotydy/dx=-2/x` ..... [1]

Now `1 + cot^2A -= csc^2A`
`:. 1 + cot^2y = 16x^4`
`:. cot^2y = 16x^4 - 1`
`:. coty = sqrt(16x^4 - 1)`

Substituting into [1] we get:
`sqrt(16x^4 - 1)dy/dx=-2/x`

`:. dy/dx=(-2/x)/sqrt(16x^4 - 1)`
`:. dy/dx=(-2)/(xsqrt(16x^4 - 1)) (=(-2)/(xsqrt((4x^2+1)(4x^2-1))))`