Question

How do you find the derivative of the function: `arccos ([2x + 1]/2)`?

Answer 1

`d/dx arccos ([2x + 1]/2) = -1/sqrt(1-(x+1/2)^2)`

Explanation:

We want the derivative of:

`arccos ([2x + 1]/2) = arccos (x+1/2)`

We can use the known result:

`d/dx arccos x = -1/sqrt(1-x^2)`

Along with the chain rule to get:

`d/dx arccos ([2x + 1]/2) = -1/sqrt(1-(x+1/2)^2) * d/dx (x+1/2)`
`" " = -1/sqrt(1-(x+1/2)^2)`