What is the derivative of 40t+t^2-1/12 t^3 ?

The derivative of 40t+t^2-1/12 t^3 is 40+2t-(t^2)/4

What is the first derivative of 40t+t^2-1/12 t^3 ?

The first derivative of 40t+t^2-1/12 t^3 is 40+2t-(t^2)/4

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derivative of

40 t + t^{2} - \frac{1}{12} t^{3}

40 + 2 t - \frac{t ^{2}}{4}

Solution steps

\frac{d}{d t} (40 t + t^{2} - \frac{1}{12} t^{3})

Simplify

40 t + t^{2} - \frac{1}{12} t^{3} : 40 t + t^{2} - \frac{t ^{3}}{12}

= \frac{d}{d t} (40 t + t^{2} - \frac{t ^{3}}{12})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (40 t) + \frac{d}{d t} (t^{2}) - \frac{d}{d t} (\frac{t ^{3}}{12})

\frac{d}{d t} (40 t) = 40

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

\frac{d}{d t} (\frac{t ^{3}}{12}) = \frac{t ^{2}}{4}

= 40 + 2 t - \frac{t ^{2}}{4}

\int \frac{10}{cos ( x ) - 1} d x

\frac{d y}{d x} = y x

f (t) = 7 sin (\frac{3 t}{2})

x \to - 1 lim (- x^{2} - 2 x + 3)

(4 x y^{2} - 3) d x + (4 x^{2} y + 4) d y = 0

What is the derivative of 40t+t^2-1/12 t^3 ?
The derivative of 40t+t^2-1/12 t^3 is 40+2t-(t^2)/4
What is the first derivative of 40t+t^2-1/12 t^3 ?
The first derivative of 40t+t^2-1/12 t^3 is 40+2t-(t^2)/4