# #6. What is the derivative of y = sin^-1(3x - 4x^3)

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### 1 Answer

The function `y = sin^-1(3x - 4x^3)` . Use the chain rule to determine the derivative `dy/dx` .

Let `z = 3x - 4x^3` , `y = sin^-1 z`

`dz/dx = 3 - 12x^2`

`dy/dz = 1/sqrt(1 - z^2)`

Substituting z = 3x - 4x^3 gives `dy/dz = 1/sqrt(1 - (3x - 4x^3)^2)`

`dy/dx = (dy/dz)*(dz/dx) = (3 - 12x^2)/sqrt(1 - (3x - 4x^3)^2)`

= `(3 - 12x^2)/sqrt((1 - 3x + 4x^3)(1 + 3x - 4x^3))`

= `(3 - 12x^2)/sqrt((1 + x - 4x - 4x^2 + 4x^2 + 4x^3)(1 - x + 4x - 4x^2 + 4x^2 - 4x^3))`

= `(3 - 12x^2)/sqrt((1 + x - 4x(1 + x) + 4x^2(1 + x))(1 - x + 4x(1 - x) + 4x^2(1 - x)))`

= `(3 - 12x^2)/sqrt((1 + x)(1 - 4x + 4x^2)(1 - x)(1 + 4x + 4x^2))`

= `(3*(1 - 4x^2))/sqrt((1 + x)(2x - 1)^2(1 - x)(2x + 1)^2)`

= `(3*(1 - 4x^2))/((2x - 1)(2x + 1)*sqrt(1 - x^2))`

= `(-3(4x^2 - 1))/((4x^2 - 1)sqrt(1 - x^2))`

`=-3/sqrt(1 - x^2)`

**The derivative of `y = sin^-1(3x - 4x^3)` is `dy/dx= -3/sqrt(1 - x^2)` **