# How do you differentiate f(x)=sqrt((1-sinx)/(1+sinx))?

Nov 3, 2016

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

#### Explanation:

Let $y = f \left(x\right) = \sqrt{\frac{1 - \sin x}{1 + \sin x}} \implies {y}^{2} = \frac{1 - \sin x}{1 + \sin x}$

The LHS can be differentiated implicit, and the RHS can be differentiated using the quotient rule:

If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$, or less formally, $\left(\frac{u}{v}\right) ' = \frac{v \left(\mathrm{du}\right) - u \left(\mathrm{dv}\right)}{v} ^ 2$

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with ${y}^{2} = \frac{1 - \sin x}{1 + \sin x}$ we have;

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

:. $2 y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 + \sin x\right) \left(- \cos x\right) - \left(1 - \sin x\right) \left(\cos x\right)}{1 + \sin x} ^ 2$
:. $2 y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \cos x - \sin x \cos x - \cos x + \sin x \cos x}{1 + \sin x} ^ 2$
:. $2 y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 \cos x}{1 + \sin x} ^ 2$
:. $y \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\cos x}{1 + \sin x} ^ 2$
:. $\left(\sqrt{\frac{1 - \sin x}{1 + \sin x}}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\cos x}{1 + \sin x} ^ 2$
:. $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\cos x}{{\left(1 + \sin x\right)}^{2} \sqrt{\frac{1 - \sin x}{1 + \sin x}}}$
:. $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\cos x}{{\left(1 + \sin x\right)}^{2}} \sqrt{\frac{1 + \sin x}{1 - \sin x}}$