Question

How do you find the derivative of Inverse trig function `y = (sin(3x) + cot(x^3))^8`?

Answer 1

`y^' = 24 * [sin(3x) + cot(x^3)]^7 * [cos(3x) - x^2csc^2(x^3)]`

Explanation:

From the looks of it, you're going to have to use the chain rule three times to differentiate this function.

Before you get started, keep in mind that

`d/dx(sinx) = cosx" "` and `" "d/dx(cotx) = -csc^2x`

So, use the chain rule first for `u^8`, with `u = sin(3x) + cot(x^3)`

`d/dx(y) = d/(du)(u^8) * d/dx(u)`

`y^' = 8u^7 * d/dx(sin(3x) + cot(x^3))`

You can differentiate this function, `d/dx(sin(3x) + cot(x^3))` by using the chain rule twice, once for `sint`, with `t = 3x`, and once more for `cotv`, with `v = x^3`.

This means that you can write

`d/dx(sint) = d/(dt)sint * d/dx(t)`

`d/dx(sint) = cost * d/dx(3x)`

`d/dx(sin(3x)) = cos(3x) * 3`

and

`d/dx(cotv) = d/(dv)cotv * d/dx(v)`

`d/dx(cotv) = -csc^2v * d/dx(x^3)`

`d/dx(cot(x^3)) = -csc^2(x^3) * 3x^2`

Plug these derivatives into the calculation for `y^'` to get

`y^' = 8u^7 * [3cos(3x) - 3x^2csc^2(x^3)]`

`y^' = 8[sin(3x) + cot(x^3)]^7 * 3 * [cos(3x) - x^2csc^2(x^3)]`

.`y^' = color(green)(24 * [sin(3x) + cot(x^3)]^7 * [cos(3x) - x^2csc^2(x^3)])`