# Find the derivative of cscx from first principles?

Jun 26, 2016

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

#### Explanation:

As $y = \csc x = \frac{1}{\sin} x$, $y + \delta y = \frac{1}{\sin} \left(x + \delta x\right)$

Hence $\delta y = \frac{1}{\sin} \left(x + \delta x\right) - \frac{1}{\sin} x$

= $\frac{\sin x - \sin \left(x + \delta x\right)}{\sin x \sin \left(x + \delta x\right)}$

= $- \frac{\sin \left(x + \delta x\right) - \sin x}{\sin x \sin \left(x + \delta x\right)}$

= $- \frac{2 \sin \left(\frac{x + \delta x - x}{2}\right) \cos \left(\frac{x + \delta x + x}{2}\right)}{\sin x \sin \left(x + \delta x\right)}$

= $- \frac{2 \sin \left(\frac{\delta x}{2}\right) \cos \left(x + \frac{\delta x}{2}\right)}{\sin x \sin \left(x + \delta x\right)}$ and

$\frac{\delta y}{\delta x} = - \frac{2 \sin \left(\frac{\delta x}{2}\right) \cos \left(x + \frac{\delta x}{2}\right)}{\delta x \sin x \sin \left(x + \delta x\right)}$

Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = L {t}_{\delta x \to 0} \frac{\delta y}{\delta x}$

= $L {t}_{\delta x \to 0} - \frac{2 \sin \left(\frac{\delta x}{2}\right) \cos \left(x + \frac{\delta x}{2}\right)}{\delta x \sin x \sin \left(x + \delta x\right)}$

= $L {t}_{\delta x \to 0} \left(- \sin \frac{\frac{\delta x}{2}}{\frac{\delta x}{2}}\right) \times L {t}_{\delta x \to 0} \left[\frac{\cos \left(x + \frac{\delta x}{2}\right)}{\delta x \sin x \sin \left(x + \delta x\right)}\right]$

= $- 1 \times \cos \frac{x}{{\sin}^{2} x} = - \frac{1}{\sin x} \times \cos \frac{x}{\sin} x = - \csc x \cot x$