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

Sep 12, 2015

${\csc}^{-} 1 \left({x}^{2} + 1\right) = \frac{- 2}{{x}^{2} + 1}$

#### Explanation:

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

Let $y = {\csc}^{-} 1 x$.
$x = \csc y$
$\frac{d}{\mathrm{dx}} x = \frac{d}{\mathrm{dx}} \csc y$

$1 = \frac{\mathrm{dy}}{\mathrm{dx}} \cdot - \csc y \cot y$

$\frac{- 1}{\csc y \cot y} = \frac{\mathrm{dy}}{\mathrm{dx}}$

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

$\frac{- 1}{\csc y \sqrt{{\csc}^{2} y - 1}} = \frac{\mathrm{dy}}{\mathrm{dx}}$

Use $x = \csc y$:

$\frac{- 1}{x \sqrt{x - 1}} = \frac{\mathrm{dy}}{\mathrm{dx}}$

Therefore $\textcolor{b l u e}{\frac{d}{\mathrm{dx}} {\csc}^{- 1} x = \frac{- 1}{x \sqrt{x - 1}}}$

To simplify $\frac{d}{\mathrm{dx}} {\csc}^{-} 1 \left({x}^{2} + 1\right)$ we can use the chain rule:
$\left(f \circ g\right) ' \left(x\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

$\frac{d}{\mathrm{dx}} {\csc}^{-} 1 \left({x}^{2} + 1\right)$
$= \frac{- 1}{\left({x}^{2} + 1\right) \sqrt{\left({x}^{2} + 1\right) - 1}} \frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right)$
$= \frac{- 1}{\left({x}^{2} + 1\right) \sqrt{{x}^{2}}} \cdot \left(2 x\right)$
$= \frac{- 2 x}{\left({x}^{2} + 1\right) x}$
$= \frac{- 2}{{x}^{2} + 1}$