# How do you differentiate y=1/sinx+1/cosx?

Oct 31, 2016

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

#### Explanation:

You should learn that $\frac{1}{\sin} x = \csc x$ and $\frac{1}{\cos} x = \sec x$
And, $\frac{d}{\mathrm{dx}} \csc x = - \csc x \cot x$ and $\frac{d}{\mathrm{dx}} \sec x = \sec x \tan x$

These resul;ts can be obtained using the quotient rule;
$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$;

$y = \frac{1}{\sin} x + \frac{1}{\cos} x$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \left\{\frac{\left(\sin x\right) \left(\frac{d}{\mathrm{dx}} 1\right) - \left(1\right) \left(\frac{d}{\mathrm{dx}} \sin x\right)}{\sin x} ^ 2\right\} + \left\{\frac{\left(\cos\right) \left(\frac{d}{\mathrm{dx}} 1\right) - \left(1\right) \left(\frac{d}{\mathrm{dx}} \cos x\right)}{\cos x} ^ 2\right\}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \left\{\frac{0 - \left(1\right) \left(\cos x\right)}{\sin x} ^ 2\right\} + \left\{\frac{0 - \left(1\right) \left(- \sin x\right)}{\cos x} ^ 2\right\}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \cos \frac{x}{\sin x} ^ 2 + \sin \frac{x}{\cos x} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \cos \frac{x}{\sin} x \frac{1}{\sin} x + \sin \frac{x}{\cos} x \frac{1}{\cos} x$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \cot x \csc x + \tan x \sec x$

Oct 31, 2016

$y = \frac{1}{\sin} x + \frac{1}{\cos} x$

$y = \csc x + \sec x$

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

To find out how to differentiate $\csc x$ using implicit differentiation watch this video:

To find out how to differentiate $\sec x$ using implicit differentiation watch this video: