Calculus Examples

$h = 40 t - 4.9 t^{2}$

Step 1

By the Sum Rule, the derivative of $40 t - 4.9 t^{2}$ with respect to $h$ is $\frac{d}{d h} [40 t] + \frac{d}{d h} [- 4.9 t^{2}]$ .

$\frac{d}{d h} [40 t] + \frac{d}{d h} [- 4.9 t^{2}]$

Step 2

Since $40 t$ is constant with respect to $h$ , the derivative of $40 t$ with respect to $h$ is $0$ .

$0 + \frac{d}{d h} [- 4.9 t^{2}]$

Step 3

Since $- 4.9 t^{2}$ is constant with respect to $h$ , the derivative of $- 4.9 t^{2}$ with respect to $h$ is $0$ .

$0 + 0$

Step 4

Add $0$ and $0$ .

$0$