What is the derivative of  arcsin(1/x)?

Jul 26, 2015

$- \frac{1}{x \sqrt{{x}^{2} - 1}}$

Explanation:

To differentiate this we will be applying a chain rule:

Start by Letting $\theta = \arcsin \left(\frac{1}{x}\right)$

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

Now differentiate each term on both sides of the equation with respect to $x$

$\implies \cos \left(\theta\right) \cdot \frac{d \left(\theta\right)}{\mathrm{dx}} = - \frac{1}{x} ^ 2$

Using the identity : ${\cos}^{2} \theta + {\sin}^{2} \theta = 1 \implies \cos \theta = \sqrt{1 - {\sin}^{2} \theta}$

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

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

Recall : $\sin \left(\theta\right) = \frac{1}{x} \text{ }$ and $\text{ } \theta = \arcsin \left(\frac{1}{x}\right)$

So we can write,

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

$= - \frac{1}{x} ^ 2 \cdot \frac{x}{\sqrt{{x}^{2} - 1}} = \textcolor{b l u e}{- \frac{1}{x \sqrt{{x}^{2} - 1}}} \text{ or } - \frac{\sqrt{{x}^{2} - 1}}{x \left({x}^{2} - 1\right)}$