# What is the derivative of arc csc(sqrt(x+1))?

Nov 13, 2015

$\frac{d}{\mathrm{dx}} \text{arccsc} \left(\sqrt{x + 1}\right) = - \frac{1}{2 \left(x + 1\right) \sqrt{x}}$

#### Explanation:

To evaluate inverse trigonometric functions, a valuable technique is implicit differentiation:
$\frac{d}{\mathrm{dx}} f \left(y\right) = \frac{d}{\mathrm{dy}} f \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

In this case, we also will use the chain rule:
$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) g ' \left(x\right)$

as well as the following derivatives:
$\frac{d}{\mathrm{dx}} \csc \left(x\right) = - \csc \left(x\right) \cot \left(x\right)$
$\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$

Now, let $y = \text{arccsc} \left(\sqrt{x + 1}\right)$
$\implies \csc \left(y\right) = \csc \left(\text{arccsc} \left(\sqrt{x + 1}\right)\right) = \sqrt{x + 1}$
$\implies \frac{d}{\mathrm{dx}} \csc \left(y\right) = \frac{d}{\mathrm{dx}} \sqrt{x + 1}$

Through implicit differentiation:
$\frac{d}{\mathrm{dx}} \csc \left(y\right) = \frac{d}{\mathrm{dy}} \csc \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - \csc \left(y\right) \cot \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

And through the chain rule
$\frac{d}{\mathrm{dx}} {\left(x + 1\right)}^{\frac{1}{2}} = \frac{1}{2} {\left(x + 1\right)}^{- \frac{1}{2}} \left(\frac{d}{\mathrm{dx}} \left(x + 1\right)\right) = \frac{1}{2 \sqrt{x + 1}}$

So
$- \csc \left(y\right) \cot \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x + 1}}$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2 \sqrt{x + 1} \cdot \csc \left(y\right) \cot \left(y\right)}$

But we want our final answer entirely in terms of $x$. To fix this, we remember that $\csc \left(y\right) = \sqrt{x + 1}$ and draw a corresponding right triangle: From this we can see that $\cot \left(y\right) = \sqrt{x}$
Thus we have
$- \frac{1}{2 \sqrt{x + 1} \cdot \csc \left(y\right) \cot \left(y\right)} = - \frac{1}{2 \sqrt{x + 1} \cdot \sqrt{x + 1} \cdot \sqrt{x}}$

From this, we can get our final answer

$\frac{d}{\mathrm{dx}} \text{arccsc} \left(\sqrt{x + 1}\right) = - \frac{1}{2 \left(x + 1\right) \sqrt{x}}$