What is expand (2x^2+6x-4)(2x^2-3x)' ?

The solution to expand (2x^2+6x-4)(2x^2-3x)' is 2x^2(2x^2-3x)'+6x(2x^2-3x)'-4(2x^2-3x)'

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expand (2x^2+6x-4)(2x^2-3x)'

expand

(2 x^{2} + 6 x - 4) (2 x^{2} - 3 x)^{'}

2 x^{2} (2 x^{2} - 3 x)' + 6 x (2 x^{2} - 3 x)' - 4 (2 x^{2} - 3 x)'

Solution steps

(2 x^{2} + 6 x - 4) (2 x^{2} - 3 x)^{'}

Distribute parentheses

= (2 x^{2} - 3 x)^{^{'}} 2 x^{2} + (2 x^{2} - 3 x)^{^{'}} 6 x + (2 x^{2} - 3 x)^{^{'}} (- 4)

Apply minus-plus rules

+ (- a) = - a

= 2 x^{2} (2 x^{2} - 3 x)^{^{'}} + 6 x (2 x^{2} - 3 x)^{^{'}} - 4 (2 x^{2} - 3 x)^{^{'}}

xx

\frac{7 x ^{2} + x + 1}{( x - 1 ) ( 2 x ^{2} + 5 x + 2 )}

\frac{( x - 4 ) ( x + 2 )}{( x - 1 ) ( x - 3 )}

sin (2 x) δ (x - 1)

2 (2 + 2 i) \frac{d}{d x} (2 + 2 i)

What is expand (2x^2+6x-4)(2x^2-3x)' ?
The solution to expand (2x^2+6x-4)(2x^2-3x)' is 2x^2(2x^2-3x)'+6x(2x^2-3x)'-4(2x^2-3x)'