Question

How do you find the derivitive of Inverse trig function `y = csc^-1(x^2+1)`?

Answer 1

`csc^-1(x^2 + 1) = (-2)/(x^2 + 1)`

Explanation:

If you don't know how to get the derivative of inverse trigonometric functions, you can obtain the formula using implicit differentiation:

Let `y = csc^-1 x`.
`x = csc y`
`d/dx x = d/dx csc y`

`1= dy/dx * -csc y cot y`

`(-1)/(csc y cot y)= dy/dx`

Use a fundamental identity to express all trig functions as the one you are differentiating, in this case: `cot^2 theta + 1 = csc^2 theta`:

`(-1)/(csc y sqrt(csc^2 y - 1) )= dy/dx`

Use `x = csc y`:

`(-1)/(x sqrt(x - 1) )= dy/dx`

Therefore `color(blue)(d/dx csc^(-1) x = (-1)/(x sqrt(x - 1) ))`

To simplify `d/dx csc^-1(x^2 + 1)` we can use the chain rule:
`(f @ g)'(x) = f'(g(x)) * g'(x)`

`d/dx csc^-1(x^2 + 1)`
`= (-1)/((x^2 + 1) sqrt((x^2 + 1)- 1) ) d/dx (x^2 + 1)`
`= (-1)/((x^2 + 1) sqrt(x^2)) * (2x)`
`= (-2x)/((x^2 + 1) x)`
`= (-2)/(x^2 + 1)`