What is the d/(dt)(1.1t+6.3t^3) ?

The d/(dt)(1.1t+6.3t^3) is 1.1+18.9t^2

What is the first d/(dt)(1.1t+6.3t^3) ?

The first d/(dt)(1.1t+6.3t^3) is 1.1+18.9t^2

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\frac{d}{d t} (1.1 t + 6.3 t^{3})

1.1 + 18.9 t^{2}

Solution steps

\frac{d}{d t} (1.1 t + 6.3 t^{3})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (1.1 t) + \frac{d}{d t} (6.3 t^{3})

\frac{d}{d t} (1.1 t) = 1.1

\frac{d}{d t} (6.3 t^{3}) = 18.9 t^{2}

= 1.1 + 18.9 t^{2}

What is the d/(dt)(1.1t+6.3t^3) ?
The d/(dt)(1.1t+6.3t^3) is 1.1+18.9t^2
What is the first d/(dt)(1.1t+6.3t^3) ?
The first d/(dt)(1.1t+6.3t^3) is 1.1+18.9t^2