Differentiating a trig fraction. Check.

Question

I am trying to differentiate this function:

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

Is this right?

$$y' = \frac{(1+\tan x)\cos x - \sin x*\sec^2x}{(1 + \tan x)^2}$$

$$= \frac{\cos x + \sin x - \frac{\sin x}{\cos^2x}}{(1+\tan x)^2}$$

Am I doing something wrong? I feel like I can simplify the $1+\tan x$ but I'm a bit unsure how...

Answer 1

Yes, your answer is correct. You can continue simplification.

First of all, you can conclude that:
$(1+\tan x)^2=(1+\frac{\sin x}{\cos x})^2=(\frac{\sin x+\cos x}{\cos x})^2=\frac{1+2\sin x \cos x}{\cos^2x}$

Then the simple form will be like below:
$y'= \frac{\cos x + \sin x - \frac{\sin x}{\cos^2x}}{(1+\tan x)^2}= \frac{(\cos^2x)(\cos x + \sin x - \frac{\sin x}{\cos^2x})}{(\cos^2x)(1+\tan x)^2}=\frac{\cos^3x+\sin x\cos^2x-\sin x}{1+2\sin x\cos x}$
$=\frac{\cos^3x-\sin x(1-\cos^2x)}{1+2\sin x \cos x}=\frac{\cos^3x-\sin^3x}{1+2\sin x\cos x}=\frac{\cos^3x-\sin^3x}{\sin 2x +1}$

Answer 2

it is by the quotient rule $$y'=\frac{\cos(x)(1+\tan(x))-\sin(x)(1+\tan^2(x))}{(1+\tan(x))^2}$$ note that $$(\tan(x))'=1+\tan^2(x)$$