What is the d/(dt)(t^3-6t^2+9t) ?

The d/(dt)(t^3-6t^2+9t) is 3t^2-12t+9

What is the first d/(dt)(t^3-6t^2+9t) ?

The first d/(dt)(t^3-6t^2+9t) is 3t^2-12t+9

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\frac{d}{d t} (t^{3} - 6 t^{2} + 9 t)

3 t^{2} - 12 t + 9

Solution steps

\frac{d}{d t} (t^{3} - 6 t^{2} + 9 t)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (t^{3}) - \frac{d}{d t} (6 t^{2}) + \frac{d}{d t} (9 t)

\frac{d}{d t} (t^{3}) = 3 t^{2}

\frac{d}{d t} (6 t^{2}) = 12 t

\frac{d}{d t} (9 t) = 9

= 3 t^{2} - 12 t + 9

\frac{d}{d t} (15 + 20 t)

\frac{( 1 + e ^{x} )}{( 1 - e ^{x} )}

y = 3 e^{x} - 14 x - 13

\frac{d}{d s} (x^{2} - 2^{s} in x)

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

What is the d/(dt)(t^3-6t^2+9t) ?
The d/(dt)(t^3-6t^2+9t) is 3t^2-12t+9
What is the first d/(dt)(t^3-6t^2+9t) ?
The first d/(dt)(t^3-6t^2+9t) is 3t^2-12t+9