# Y=cosx+sinx/cosx-sinx find dy/dx?

May 15, 2018

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

#### Explanation:

show below

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

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

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

$y ' = 1 - \frac{\left(- \sin \left(x\right) - \cos \left(x\right)\right) \cdot \left(\sin \left(x\right) + \cos \left(x\right)\right)}{\cos \left(x\right) - \sin \left(x\right)} ^ 2$

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

May 15, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} \left(\frac{\pi}{4} + x\right) = \frac{2}{\cos x - \sin x} ^ 2 = \frac{2}{1 - \sin 2 x}$.

#### Explanation:

I presume that, $y = \frac{\cos x + \sin x}{\cos x - \sin x}$,

$= \frac{\cos x \left(1 + \sin \frac{x}{\cos} x\right)}{\cos x \left(1 - \sin \frac{x}{\cos} x\right)}$,

$= \frac{1 + \tan x}{1 - \tan x}$,

$\Rightarrow y = \tan \left(\frac{\pi}{4} + x\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} \left(\frac{\pi}{4} + x\right) \cdot \frac{d}{\mathrm{dx}} \left(\frac{\pi}{4} + x\right) \ldots \text{[The Chain Rule]}$,

$= {\sec}^{2} \left(\frac{\pi}{4} + x\right)$.

Also, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} ^ 2 \left(\frac{\pi}{4} + x\right)$,

$= \frac{1}{\cos \left(\frac{\pi}{4} + x\right)} ^ 2$,

$= \frac{1}{\cos \left(\frac{\pi}{4}\right) \cos x - \sin \left(\frac{\pi}{4}\right) \sin x} ^ 2$,

$= \frac{1}{\frac{1}{\sqrt{2}} \cdot \cos x - \frac{1}{\sqrt{2}} \cdot \sin x} ^ 2$.

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{\cos x - \sin x} ^ 2 , \mathmr{and} ,$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{{\cos}^{2} x - 2 \cos x \sin x + {\sin}^{2} x} = \frac{2}{1 - \sin 2 x}$.