What is the d/(dt)(at^5-5bt^3) ?

The d/(dt)(at^5-5bt^3) is 5at^4-15bt^2

What is the first d/(dt)(at^5-5bt^3) ?

The first d/(dt)(at^5-5bt^3) is 5at^4-15bt^2

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\frac{d}{d t} (a t^{5} - 5 b t^{3})

5 a t^{4} - 15 b t^{2}

Solution steps

\frac{d}{d t} (a t^{5} - 5 b t^{3})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (a t^{5}) - \frac{d}{d t} (5 b t^{3})

\frac{d}{d t} (a t^{5}) = 5 a t^{4}

\frac{d}{d t} (5 b t^{3}) = 15 b t^{2}

= 5 a t^{4} - 15 b t^{2}

What is the d/(dt)(at^5-5bt^3) ?
The d/(dt)(at^5-5bt^3) is 5at^4-15bt^2
What is the first d/(dt)(at^5-5bt^3) ?
The first d/(dt)(at^5-5bt^3) is 5at^4-15bt^2