Question

What is the derivative of this function `y=x^2+cos^-1x`?

Answer 1

`y' = 2x - 1/sinx`

Explanation:

`y = x^2 + cos^-1 x`
`y = x^2 + 1/cosx`
`d/dx y = d/dx (x^2 + 1/cosx)`
`y' = 2x + (1/-sinx)`
`y' = 2x - 1/sinx`

you have to remember the trig table of derivatives
`d/dx sinx = cosx`
`d/dx cosx = -sinx`