Calculus Examples

$f (x) = 0.0000329 t^{3} - 0.0061 t^{2} + 0.0514 t + 417$

Step 1

Find the first derivative.

Tap for more steps...

Step 1.1

Find the first derivative.

Tap for more steps...

Step 1.1.1

By the Sum Rule, the derivative of

$0.0000329 t^{3} - 0.0061 t^{2} + 0.0514 t + 417$ with respect to

$x$ is

$\frac{d}{d x} [0.0000329 t^{3}] + \frac{d}{d x} [- 0.0061 t^{2}] + \frac{d}{d x} [0.0514 t] + \frac{d}{d x} [417]$ .

$\frac{d}{d x} [0.0000329 t^{3}] + \frac{d}{d x} [- 0.0061 t^{2}] + \frac{d}{d x} [0.0514 t] + \frac{d}{d x} [417]$

Step 1.1.2

Since

$0.0000329 t^{3}$ is constant with respect to

$x$ , the derivative of

$0.0000329 t^{3}$ with respect to

$x$ is

$0$ .

$0 + \frac{d}{d x} [- 0.0061 t^{2}] + \frac{d}{d x} [0.0514 t] + \frac{d}{d x} [417]$

Step 1.1.3

Since

$- 0.0061 t^{2}$ is constant with respect to

$x$ , the derivative of

$- 0.0061 t^{2}$ with respect to

$x$ is

$0$ .

$0 + 0 + \frac{d}{d x} [0.0514 t] + \frac{d}{d x} [417]$

Step 1.1.4

Since

$0.0514 t$ is constant with respect to

$x$ , the derivative of

$0.0514 t$ with respect to

$x$ is

$0$ .

$0 + 0 + 0 + \frac{d}{d x} [417]$

Step 1.1.5

Since

$417$ is constant with respect to

$x$ , the derivative of

$417$ with respect to

$x$ is

$0$ .

$0 + 0 + 0 + 0$

Step 1.1.6

Combine terms.

Tap for more steps...

Step 1.1.6.1

Add

$0$ and

$0$ .

$0 + 0 + 0$

Step 1.1.6.2

Add

$0$ and

$0$ .

$0 + 0$

Step 1.1.6.3

Add

$0$ and

$0$ .

$f' (x) = 0$

Step 1.2

The first derivative of

$f (x)$ with respect to

$x$ is

$0$ .

$0$

Step 2

Set the first derivative equal to

$0$ then solve the equation

$0 = 0$ .

Tap for more steps...

Step 2.1

Set the first derivative equal to

$0$ .

$0 = 0$

Step 2.2

Since

$0 = 0$ , the equation will always be true.

Always true

Step 3

There are no values of

$x$ in the domain of the original problem where the derivative is

$0$ or undefined.

No critical points found