Differentiate $\sqrt{\frac{1 +\sin x}{1 -\sin x}}$ I have tried a lot of ways to solve this question but I am unable to get the answer as same as my textbook. 
The text book answer is as follow: $$\frac{1}{2}\sec^2\left(\frac{\pi}{4}+\frac{{x}}{2}\right)$$
The steps which I took is as follows:
$$\sqrt{\frac{1+ \sin x}{1-\sin x}\cdot \frac{1+\sin x}{1+\sin x}}$$
Then Secondly
$$\sqrt{\frac{\left(1+\sin x\right)^2}{1-\sin^2 x}}$$
Then I got 
$$\dfrac{1+\sin x}{\cos x}$$
When I differentiated this I got the following 
$$\frac{\cos ^2\left(x\right)+\sin \left(x\right)\left(1+\sin \left(x\right)\right)}{\cos ^2\left(x\right)}$$
Can Anyone tell me what I am doing wrong?
I also know that $$\sec^2\left(\frac{\pi}{4}+\frac{{x}}{2}\right)=\frac{2}{\left(\cos  \frac{x}{2}-\sin\frac{x}{2}\right)^2}$$
Thank you for the help!
 A: Why make things complicated if there is a easy way? By the half-angle formulae we obtain
$$\sqrt{\frac{1-\sin(x)}{1+\sin(x)}}=\sqrt{\frac{1-\cos\left(x+\frac\pi2\right)}{1+\cos\left(x+\frac\pi2\right)}}=\tan\left(\frac x2-\frac\pi4\right)$$
And I suppose you can differentiate the tangent function ;) 

As pointed out by Simply Beautiful Art and mathcounterexamples.net by using the half-angle formula we ran into serious issus concerning the sign.
A: Logarithmic differentiation makes things easier
$$y=\sqrt{\dfrac{1 +\sin (x)}{1 -\sin (x)}}\implies \log(y)=\frac 12 \left(\log(1+\sin(x)) -\log(1-\sin(x))\right)$$
$$\frac {y'}{y}=\frac 12 \left(\frac{\cos(x)}{1+\sin(x) }+\frac{\cos(x)}{1-\sin(x) }\right)$$ Simplify as much as you can and, when finished, use
$$y'=y\times \frac {y'}{y}$$
A: Alternatively, using the product rule:
$$\begin{align}\left(\sqrt{\dfrac{1 +\sin x}{1 -\sin x}}\right)'
&=(\sqrt{1+\sin x})'\cdot (1-\sin x)^{-1/2}+\sqrt{1+\sin x}\cdot ((1-\sin x)^{-1/2})'=\\
&=\frac{\cos x}{2\sqrt{1+\sin x}}\cdot \frac1{\sqrt{1-\sin x}}+\sqrt{1+\sin x}\cdot \frac{\cos x}{2(1-\sin x)\sqrt{1-\sin x}}=\\
&=\frac{\cos x}{2\sqrt{\cos ^2x}}+\frac{\cos x\sqrt{(1+\sin x)^2}}{2(1-\sin x)\sqrt{1-\sin ^2x}}=\\
&=\frac12+\frac{1+\sin x}{2(1-\sin x)}=\\
&=\frac1{1-\sin x}=\cdots =\\
&=\frac{1}{2}\sec^2\left(\frac{\pi}{4}+\frac{{x}}{2}\right)\end{align}$$
Can you show the equality of the last two expressions using what you stated you know?
Answer (see the hidden area):

$$\frac1{1-\sin x}=\frac{1}{\sin^2x+\cos^2x-2\sin \frac{x}{2}\cos \frac{x}{2}}=\frac{1}{(\sin \frac x2-\cos \frac x2)^2}=\\=\frac{1}{2(\frac{1}{\sqrt{2}}\sin \frac x2-\frac{1}{\sqrt{2}}\cos \frac x2)^2}=\frac{1}{2\cos^2(\frac{\pi}{4}+\frac x2)}=\frac12\sec^2(\frac{\pi}{4}+\frac x2).$$

A: What if $1\pm\sin x=0?$
Otherwise
$$\sqrt{\dfrac{1+\sin x}{1-\sin x}}=\sqrt{\left(\dfrac{1+\tan\dfrac x2}{1-\tan\dfrac x2}\right)^2}=\left|\tan\left(\dfrac\pi4+\dfrac x2\right)\right|$$ using
https://www.cut-the-knot.org/arithmetic/algebra/WeierstrassSubstitution.shtml
Now $\tan\left(\dfrac\pi4+\dfrac x2\right)$ will be $>0$ if $1-\tan^2\dfrac x2>0\iff-1<\tan\dfrac x2<1$
A: The textbook answer may not be quite right.
For x in range $[0, 2\pi],\text{ only } [0, {\pi \over 2}], [{3\pi \over 2},2\pi]$ work.  
This is a revised derivative, to handle the full range.
$$\left(\sqrt{{1+\sin(x) \over 1-\sin(x)}} \right)' = {sign(\cos(x))\over 2} \sec^2({\pi \over 4} + {x \over 2})$$
Of course, chain rule result works too, but a bit messy.
After some simplification, this is what I have.
Note: derivative have the sign of $\cos(x)$, as expected.
$$\left(\sqrt{{1+\sin(x) \over 1-\sin(x)}} \right)'
= {\cos(x) \over (1-\sin(x))^2 \sqrt{{1+sin(x) \over 1-sin(x)}}}$$
