We start with the chain rule:
y ′ = − sin ( x 2 + x 2 + 1 ) ⋅ d d x { x 2 + x 2 + 1 } = − sin ( x 2 + x 2 + 1 ) ⋅ ( 2 x + d d x { x 2 + 1 } ) and find ourselves in need of chain rule a second time: = − sin ( x 2 + x 2 + 1 ) ⋅ ( 2 x + 1 2 x 2 + 1 ⋅ d d x { x 2 + 1 } ) = − sin ( x 2 + x 2 + 1 ) ⋅ ( 2 x + 2 x 2 x 2 + 1 ) \begin{aligned}
y'&=-\sin\big( x^2+\sqrt{x^2+1}\big)\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left\{ x^2+\sqrt{x^2+1}\right\}\\
&=-\sin\big(x^2+\sqrt{x^2+1}\big)\cdot\left(2x+\frac{\mathrm{d}}{\mathrm{d}x}\left\{\sqrt{x^2+1}\right\}\right)
\\ \text{and find ourselves in need of chain rule a second time:} \\
&=-\sin\big(x^2+\sqrt{x^2+1}\big)\cdot\left(2x+\dfrac{1}{2\sqrt{ x^2+1}}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left\{ x^2+1\right\}\right)\\
&=-\sin\big(x^2+\sqrt{x^2+1}\big)\cdot\left(2x+\dfrac{2x}{2\sqrt{x^2+1}}\right)
\end{aligned} y ′ and find ourselves in need of chain rule a second time: = − sin ( x 2 + x 2 + 1 ) ⋅ d x d { x 2 + x 2 + 1 } = − sin ( x 2 + x 2 + 1 ) ⋅ ( 2 x + d x d { x 2 + 1 } ) = − sin ( x 2 + x 2 + 1 ) ⋅ ( 2 x + 2 x 2 + 1 1 ⋅ d x d { x 2 + 1 } ) = − sin ( x 2 + x 2 + 1 ) ⋅ ( 2 x + 2 x 2 + 1 2 x )