# How do you find the derivative of y=arcsin(1/x)?

Jul 31, 2014

First, recall the identity $\frac{d}{\mathrm{dx}} \left[\arcsin \alpha\right] = \frac{1}{\sqrt{1 - {x}^{2}}}$.

If this identity doesn't look familiar then I may recommend viewing a few videos from this page as they present a couple identities like this, and explain why they are true.

Differentiating $\arcsin \left(\frac{1}{x}\right)$ is just a matter of using the identity above, as well as the chain rule:

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

The derivative of $\frac{1}{x}$ is found using the power rule:

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

Now, all we need to do is simplify a bit:

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{{x}^{2} / \left\mid x \right\mid \sqrt{{x}^{2} - 1}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\left\mid x \right\mid \sqrt{{x}^{2} - 1}}$

Oct 2, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\left\mid x \right\mid \sqrt{{x}^{2} - 1}}$

#### Explanation:

We may also know from the outset that:

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

Then:

$y = \arcsin \left(\frac{1}{x}\right)$

$\sin \left(y\right) = \frac{1}{x}$

$\frac{1}{\sin \left(y\right)} = x$

$\csc \left(y\right) = x$

$y = \text{arccsc} \left(x\right)$

Then:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\left\mid x \right\mid \sqrt{{x}^{2} - 1}}$