The given function is \( \displaystyle y = \left( \frac{2\tan x}{\tan x + \cos x} \right)^2 \) and we need to find the derivative of this function with respect to \( x \).
Solution:
Let \( \displaystyle u = \frac{2\tan x}{\tan x + \cos x} \), so \( y = u^2 \).
Using the chain rule:
\(\displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 2u \cdot \frac{du}{dx} \)
Now differentiate \( u \) with respect to \( x \).
\( \begin{aligned} \frac{du}{dx} &= \frac{ (\tan x + \cos x) \cdot \frac{d}{dx}(2\tan x) \,- (2\tan x) \cdot \frac{d}{dx}(\tan x + \cos x) }{(\tan x + \cos x)^2} \\ &= \frac{(\tan x + \cos x)(2\sec^2 x) \,- (2\tan x)(\sec^2 x \,- \sin x)}{(\tan x + \cos x)^2} \\ &= \frac{2\sec^2 x \tan x + 2\sec^2 x \cos x \,- 2\sec^2 x \tan x + 2\tan x \sin x}{(\tan x + \cos x)^2} \\ &= \frac{2\sec^2 x \cos x + 2\tan x \sin x}{(\tan x + \cos x)^2} \\ &= \frac{2\sec x + 2\tan x \sin x}{(\tan x + \cos x)^2} \\ \end{aligned} \)Substitute back into \( \frac{dy}{dx} \):
\( \begin{aligned} \frac{dy}{dx} &= 2u \cdot \frac{du}{dx} = 2 \left( \frac{2\tan x}{\tan x + \cos x} \right) \cdot \frac{2\sec x + 2\tan x \sin x}{(\tan x + \cos x)^2} \\ &= \frac{8\tan x (\sec x + \tan x \sin x)}{(\tan x + \cos x)^3} \\ \end{aligned} \)Thus, the derivative of the function is:
\( \boxed{ \frac{dy}{dx} = \frac{8\tan x (\sec x + \tan x \sin x)}{(\tan x + \cos x)^3} } \)Please let me know in the comments if you find any errors in this solution.