Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 46

Determine the derivative of the function $y = \left[ x + ( x + \sin^2x)^3\right]^4$


$\begin{equation} \begin{aligned} y' &= \frac{d}{dx} \left[ x + ( x + \sin^2x)^3\right]^4\\ \\ y' &= 4 \left[ x + ( x + \sin^2x)^3\right]^3 \frac{d}{dx} \left[ x + ( x + \sin^2x)^3\right]\\ \\ y' &= 4 \left[ x + ( x + \sin^2x)^3\right]^3 \left[ \frac{d}{dx} (x) + \frac{d}{dx} ( x + \sin^2 x)^3 \right]\\ \\ y' &= 4 \left[ x + ( x + \sin^2x)^3\right]^3 \left[ 1+3(x+\sin^2x)^2 \frac{d}{dx} (x+\sin^2x)\right]\\ \\ y' &= 4 \left[ x + ( x + \sin^2x)^3\right]^3 \left[ 1+3(x+\sin^2x)^2 \left(\frac{d}{dx} (x) + \frac{d}{dx} (\sin x)^2 \right)\right]\\ \\ y' &= 4 \left[ x + ( x + \sin^2x)^3\right]^3 \left[ 1 +3 ( x+ \sin^2x)^2 \left(1+2(\sin x)\frac{d}{dx}(\sin x)\right)\right]\\ \\ y' &= 4 \left[ x + ( x + \sin^2x)^3\right]^3 \left[ 1+3 (x + \sin^2x)^2(1+2 \sin x \cos x) \right] && \left( \text{Recall the Double Angle Formula } (\sin 2x = 2 \sin x \cos x) \right)\\ \\ y' &= 4 \left[ x + ( x + \sin^2x)^3\right]^3 \left[ 1+3 (x + \sin^2 x)^2 ( 1+ \sin 2x)\right] \end{aligned} \end{equation}$