# Find derivative of cscx using first principal or definition of derivative?

Feb 11, 2018

$\frac{d}{\mathrm{dx}} \left(\csc x\right) = - \cot x \csc x$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left(\csc x\right)$

$= \frac{d}{\mathrm{dx}} \left(\frac{1}{\sin x}\right)$

Using the limit definition of the derivative:
Let $f \left(x\right) = \csc x = \frac{1}{\sin} x$

$= {\lim}_{h \to 0} \frac{f \left(x + h\right) - h \left(x\right)}{h}$

$= {\lim}_{h \to 0} \frac{\csc \left(x + h\right) - \csc x}{h}$

$= {\lim}_{h \to 0} \frac{\left(\frac{1}{\sin} \left(x + h\right) - \frac{1}{\sin} x\right)}{h}$

= ${\lim}_{h \to 0} \frac{\sin x - \sin \left(x + h\right)}{h \sin \left(x + h\right) \sin x}$

= ${\lim}_{h \to 0} \frac{2 \cos \left(\frac{x + x + h}{2}\right) \sin \left(\frac{x - x - h}{2}\right)}{h \sin \left(x + h\right) \sin x}$

= ${\lim}_{h \to 0} \frac{- 2 \cos \left(x + \frac{h}{2}\right) \sin \left(\frac{h}{2}\right)}{h \sin \left(x + h\right) \sin x}$

= ${\lim}_{h \to 0} \frac{- 2 \cos \left(x + \frac{h}{2}\right)}{h \sin \left(x + h\right) \sin x} \times \sin \frac{\frac{h}{2}}{\frac{h}{2}} \times \frac{h}{2}$

= ${\lim}_{h \to 0} \frac{- \cos \left(x + \frac{h}{2}\right)}{\sin \left(x + h\right) \sin x} \times \sin \frac{\frac{h}{2}}{\frac{h}{2}}$

= ${\lim}_{h \to 0} \frac{- \cos \left(x + \frac{h}{2}\right)}{\sin \left(x + h\right) \sin x} \times {\lim}_{h \to 0} \sin \frac{\frac{h}{2}}{\frac{h}{2}}$

= $- \cos \frac{x}{\sin x \sin x} \times 1$

= $- \cot x \csc x$