Calculus Examples

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

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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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.

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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 $0 = 0$ .

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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