The given function is \(\displaystyle y = \frac{1}{\left(1+\tan ^{3} x\right)^{2}} \) and we need to find the derivative of this function with respect to \( x \).
Solution:
To find the derivative of the function \(\displaystyle y = \frac{1}{\left(1+\tan ^{3} x\right)^{2}} \) with respect to \( x \), we apply the chain rule.
\( \displaystyle \frac{dy}{dx} = \frac{d}{dx} \left[\frac{1}{\left(1 + \tan^3 x\right)^2} \right] \\ \displaystyle = \frac{d}{dx} \left[\left(1 + \tan^3 x\right)^{-2} \right] \\ \displaystyle = \frac{d\left[\left(1+\tan^{3} x\right)^{-2}\right]}{d[1+\tan ^3 x]} \cdot \frac{d\left(1+\tan ^{3} x\right)}{d x} \\ \displaystyle = -2\left(1+\tan ^{3} x\right)^{-3} \cdot\left[\frac{d(1)}{dx}+\frac{d\left(\tan ^{3} x\right)}{d(\tan x)} \cdot \frac{d(\tan x)}{dx}\right] \\ \displaystyle = \frac{-2}{\left(1+\tan ^{3} x\right)^{3}} \cdot 3 \tan ^{2} x \cdot \sec ^{2} x \\ \displaystyle = -\frac{6 \tan^2 x \cdot \sec^2 x}{\left(1 + \tan^3 x\right)^3} \\ \)Thus, the derivative of the function is:
\( \boxed{\frac{dy}{dx} = -\frac{6 \tan^2 x \cdot \sec^2 x}{\left(1 + \tan^3 x\right)^3} } \\ \)Please let me know in the comments if you find any errors in this solution.