What is the derivative of ((t-4)^3)/6+t^2 ?

The derivative of ((t-4)^3)/6+t^2 is (t^2-4t+16)/2

What is the first derivative of ((t-4)^3)/6+t^2 ?

The first derivative of ((t-4)^3)/6+t^2 is (t^2-4t+16)/2

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derivative of

\frac{( t - 4 ) ^{3}}{6} + t^{2}

\frac{t ^{2} - 4 t + 16}{2}

Solution steps

\frac{d}{d t} (\frac{( t - 4 ) ^{3}}{6} + t^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (\frac{( t - 4 ) ^{3}}{6}) + \frac{d}{d t} (t^{2})

\frac{d}{d t} (\frac{( t - 4 ) ^{3}}{6}) = \frac{( t - 4 ) ^{2}}{2}

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

= \frac{( t - 4 ) ^{2}}{2} + 2 t

Simplify

\frac{( t - 4 ) ^{2}}{2} + 2 t : \frac{t ^{2} - 4 t + 16}{2}

= \frac{t ^{2} - 4 t + 16}{2}

x \to 0 lim (9 sin (x) ln (x))

\int x ln (y) d y

x \to \frac{1}{2} lim (\frac{x}{2 x - 1})

f (x) = (x^{3} + 3 x) (2 x + 4)

\int (e^{x}) \cdot cos (n x) d x

What is the derivative of ((t-4)^3)/6+t^2 ?
The derivative of ((t-4)^3)/6+t^2 is (t^2-4t+16)/2
What is the first derivative of ((t-4)^3)/6+t^2 ?
The first derivative of ((t-4)^3)/6+t^2 is (t^2-4t+16)/2