What is the first d/(dt)(2+1t+5t^2) ?

The first d/(dt)(2+1t+5t^2) is 10t+1

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d/(dt)(2+1t+5t^2)

\frac{d}{d t} (2 + 1 t + 5 t^{2})

10 t + 1

Solution steps

\frac{d}{d t} (2 + 1 \cdot t + 5 t^{2})

Simplify

2 + 1 \cdot t + 5 t^{2} : 2 + t + 5 t^{2}

= \frac{d}{d t} (2 + t + 5 t^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (2) + \frac{d t}{d t} + \frac{d}{d t} (5 t^{2})

\frac{d}{d t} (2) = 0

\frac{d t}{d t} = 1

\frac{d}{d t} (5 t^{2}) = 10 t

= 0 + 1 + 10 t

Simplify

= 10 t + 1

\frac{x ^{2}}{( x ^{3} + 1 )}

x^{3} - x - 1

cos^{4} (\frac{x}{2}) - sin^{4} (\frac{x}{2})

\frac{- x ^{3} + x - 1}{2}

y = (x^{3} - x) e^{- 3} x