What is expand 4y^2*(4yy'(t)-1) ?

The solution to expand 4y^2*(4yy'(t)-1) is 16y^3y'(t)-4y^2

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expand 4y^2*(4yy'(t)-1)

expand

4 y^{2} \cdot (4 y y^{'} (t) - 1)

16 y^{3} y' (t) - 4 y^{2}

Solution steps

4 y^{2} (4 y y^{'} (t) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 4 y^{2}, b = 4 y y^{'} (t), c = 1

= 4 y^{2} \cdot 4 y y^{^{'}} (t) - 4 y^{2} \cdot 1

= 4 \cdot 4 y^{2} y y^{^{'}} (t) - 4 \cdot 1 \cdot y^{2}

Simplify

4 \cdot 4 y^{2} y y^{'} (t) - 4 \cdot 1 \cdot y^{2} : 16 y^{3} y^{'} (t) - 4 y^{2}

= 16 y^{3} y^{'} (t) - 4 y^{2}

(\frac{2}{3} a - \frac{1}{3} b)^{2}

(x - \frac{1}{a}) (x - \frac{1}{b}) (x - \frac{1}{c})

\frac{( x ^{2} - 2 x + 3 ) ( 2 x + 3 )}{( x - 2 ) ( x + 5 )}

so l v e (30 - 2 h) (30 - h) (h)

(3 x) l e f t (- 118.33 x - 10 x^{2})

What is expand 4y^2*(4yy'(t)-1) ?
The solution to expand 4y^2*(4yy'(t)-1) is 16y^3y'(t)-4y^2