What is the first (6x-0.75x^2)^' ?

The first (6x-0.75x^2)^' is 6-1.5x

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(6x-0.75x^2)^'

(6 x - 0.75 x^{2})^{^{'}}

6 - 1.5 x

Solution steps

(6 x - 0.75 x^{2})^{'}

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= (6 x)^{^{'}} - (0.75 x^{2})^{^{'}}

(6 x)^{'} = 6

(0.75 x^{2})^{'} = 1.5 x

= 6 - 1.5 x

\frac{d}{d x} (x (x - 2)^{3})

\int \frac{y ^{2}}{y ^{6} + 1} d y

n = 1 \sum \infty 7 (- 4)^{n - 1}

\frac{1}{( x ^{2} + 1 )}

5 x, x^{2}