Question

How do you find the derivative of `y=arcsin(x^4)`?

Answer 1

`dy/dx = (4x^3)/sqrt(1-x^8)`

Explanation:

When tackling the derivative of inverse trig functions. I prefer to rearrange and use Implicit differentiation as I always get the inverse derivatives muddled up, and this way I do not need to remember the inverse derivatives. If you can remember the inverse derivatives then you can use the chain rule.

`y=arcsin(x^4) <=> siny=x^4`

Differentiate Implicitly:

`cosydy/dx = 4x^3` ..... [1]

Using the `sin"/"cos` identity;

`sin^2y+cos^2y -= csc^2y`
`:. (x^4)^2+cos^2y=1`
`:. cos^2y=1-x^8`
`:. cosy=sqrt(1-x^8)`

Substituting into [1]
`:. sqrt(1-x^8)dy/dx=4x^3`
`:. dy/dx = (4x^3)/sqrt(1-x^8)`