find $\frac{dy}{dx}$ for $y = \frac{\tan x}{1+\tan x}$\n$\frac{dy}{dx}=square$

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $\frac{dy}{dx}=\frac{u'v - uv'}{v^{2}}$. Here, $u = \tan x$, $u'=\sec^{2}x$, $v = 1+\tan x$, and $v'=\sec^{2}x$.

Step2: Substitute values into the quotient - rule

$\frac{dy}{dx}=\frac{\sec^{2}x(1 + \tan x)-\tan x\sec^{2}x}{(1 + \tan x)^{2}}$.

Step3: Simplify the numerator

Expand the numerator: $\sec^{2}x(1 + \tan x)-\tan x\sec^{2}x=\sec^{2}x+\sec^{2}x\tan x-\sec^{2}x\tan x=\sec^{2}x$.

Answer:

$\frac{\sec^{2}x}{(1 + \tan x)^{2}}$