Find dh/dt h=-16t^2+368t

$h = - 16 t^{2} + 368 t$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d t} (h) = \frac{d}{d t} (- 16 t^{2} + 368 t)$

Step 2

The derivative of $h$ with respect to $t$ is $h'$ .

$h'$

Step 3

Differentiate the right side of the equation.

By the Sum Rule, the derivative of $- 16 t^{2} + 368 t$ with respect to $t$ is $\frac{d}{d t} [- 16 t^{2}] + \frac{d}{d t} [368 t]$ .

$\frac{d}{d t} [- 16 t^{2}] + \frac{d}{d t} [368 t]$

Evaluate $\frac{d}{d t} [- 16 t^{2}]$ .

Since $- 16$ is constant with respect to $t$ , the derivative of $- 16 t^{2}$ with respect to $t$ is $- 16 \frac{d}{d t} [t^{2}]$ .

$- 16 \frac{d}{d t} [t^{2}] + \frac{d}{d t} [368 t]$

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 2$ .

$- 16 (2 t) + \frac{d}{d t} [368 t]$

Multiply $2$ by $- 16$ .

$- 32 t + \frac{d}{d t} [368 t]$

Evaluate $\frac{d}{d t} [368 t]$ .

Since $368$ is constant with respect to $t$ , the derivative of $368 t$ with respect to $t$ is $368 \frac{d}{d t} [t]$ .

$- 32 t + 368 \frac{d}{d t} [t]$

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$- 32 t + 368 \cdot 1$

Multiply $368$ by $1$ .

$- 32 t + 368$

Step 4

Reform the equation by setting the left side equal to the right side.

$h' = - 32 t + 368$

Step 5

Replace $h'$ with $\frac{d h}{d t}$ .

$\frac{d h}{d t} = - 32 t + 368$