What is expand (y''(t)+2y'(t)+2)^2 ?

The solution to expand (y''(t)+2y'(t)+2)^2 is 4y'(t)y''(t)+4(y'(t))^2+8y'(t)+(y''(t))^2+4y''(t)+4

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expand (y''(t)+2y'(t)+2)^2

expand

(y^{''} (t) + 2 y^{'} (t) + 2)^{2}

4 y' (t) y'' (t) + 4 (y' (t))^{2} + 8 y' (t) + (y'' (t))^{2} + 4 y'' (t) + 4

Solution steps

(y^{''} (t) + 2 y^{'} (t) + 2)^{2}

(y^{''} (t) + 2 y^{'} (t) + 2)^{2} = (y^{''} (t) + 2 y^{'} (t) + 2) (y^{''} (t) + 2 y^{'} (t) + 2)

= (y^{^{''}} (t) + 2 y^{^{'}} (t) + 2) (y^{^{''}} (t) + 2 y^{^{'}} (t) + 2)

Expand

(y^{''} (t) + 2 y^{'} (t) + 2) (y^{''} (t) + 2 y^{'} (t) + 2) : 4 y^{'} (t) y^{''} (t) + 4 (y^{'} (t))^{2} + 8 y^{'} (t) + (y^{''} (t))^{2} + 4 y^{''} (t) + 4

= 4 y^{'} (t) y^{''} (t) + 4 (y^{'} (t))^{2} + 8 y^{'} (t) + (y^{''} (t))^{2} + 4 y^{''} (t) + 4

cos (θ) cot (θ) (sec (θ) - 2 tan (θ))

tan (x) (1 - sin (2 x)) sec (x)

\frac{3 x}{sin ( x ) ( x - 3 )}

\frac{1}{4 ( 8 x + 12 )}

2 (\frac{y}{3} + 1)

What is expand (y''(t)+2y'(t)+2)^2 ?
The solution to expand (y''(t)+2y'(t)+2)^2 is 4y'(t)y''(t)+4(y'(t))^2+8y'(t)+(y''(t))^2+4y''(t)+4