derivative of ((t-4)^3)/6+t^2 — step-by-step solution

Let's differentiate:

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

ⓘ The derivative of a sum is the sum of the derivatives:

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

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

We differentiate using:

\frac{d}{d t} (f^{n}) = n \cdot \frac{d}{d t} (f) \cdot f^{n - 1}

(where f =

t - 4

)

ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u'. The derivative of x^n is n*x^(n-1):

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

\frac{1}{6} \cdot 3 \cdot \frac{d}{d t} (t - 4) \cdot {(t - 4)}^{2}

■The derivative of x^n is n*x^(n-1):

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

2 t

\frac{1}{6} \cdot 3 \cdot \frac{d}{d t} (t - 4) \cdot {(t - 4)}^{2} + 2 t

ⓘ Multiplications are done before additions and subtractions:

\frac{1}{6} \cdot 3

\frac{1}{2}

■The derivative of a sum is the sum of the derivatives:

\frac{d}{d t} (t - 4)

1 + 0

\frac{1}{2} \cdot (1 + 0) \cdot {(t - 4)}^{2} + 2 t

ⓘ Adding 0 changes nothing:

1 + 0

1

\frac{1}{2} \cdot {(t - 4)}^{2} + 2 t

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{1}{2} \cdot {(t - 4)}^{2}

\frac{1 \cdot {(t - 4)}^{2}}{2}

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

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

▶Derivative:

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