Calculus Examples

$f (x) = 3 x^{5} - x^{3} + 4 x - 2$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $3 x^{5} - x^{3} + 4 x - 2$ with respect to $x$ is $\frac{d}{d x} [3 x^{5}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [- 2]$ .

$\frac{d}{d x} [3 x^{5}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [- 2]$

Step 1.2

Evaluate $\frac{d}{d x} [3 x^{5}]$ .

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Step 1.2.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{5}$ with respect to $x$ is $3 \frac{d}{d x} [x^{5}]$ .

$3 \frac{d}{d x} [x^{5}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [- 2]$

Step 1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$3 (5 x^{4}) + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [- 2]$

Step 1.2.3

Multiply $5$ by $3$ .

$15 x^{4} + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [- 2]$

Step 1.3

Evaluate $\frac{d}{d x} [- x^{3}]$ .

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Step 1.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{3}$ with respect to $x$ is $- \frac{d}{d x} [x^{3}]$ .

$15 x^{4} - \frac{d}{d x} [x^{3}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [- 2]$

Step 1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$15 x^{4} - (3 x^{2}) + \frac{d}{d x} [4 x] + \frac{d}{d x} [- 2]$

Step 1.3.3

Multiply $3$ by $- 1$ .

$15 x^{4} - 3 x^{2} + \frac{d}{d x} [4 x] + \frac{d}{d x} [- 2]$

Step 1.4

Evaluate $\frac{d}{d x} [4 x]$ .

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Step 1.4.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$15 x^{4} - 3 x^{2} + 4 \frac{d}{d x} [x] + \frac{d}{d x} [- 2]$

Step 1.4.2

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

$15 x^{4} - 3 x^{2} + 4 \cdot 1 + \frac{d}{d x} [- 2]$

Step 1.4.3

Multiply $4$ by $1$ .

$15 x^{4} - 3 x^{2} + 4 + \frac{d}{d x} [- 2]$

Step 1.5

Differentiate using the Constant Rule.

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Step 1.5.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$15 x^{4} - 3 x^{2} + 4 + 0$

Step 1.5.2

Add $15 x^{4} - 3 x^{2} + 4$ and $0$ .

$15 x^{4} - 3 x^{2} + 4$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $15 x^{4} - 3 x^{2} + 4$ with respect to $x$ is $\frac{d}{d x} [15 x^{4}] + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [4]$ .

$f'' (x) = \frac{d}{d x} (15 x^{4}) + \frac{d}{d x} (- 3 x^{2}) + \frac{d}{d x} (4)$

Step 2.2

Evaluate $\frac{d}{d x} [15 x^{4}]$ .

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Step 2.2.1

Since $15$ is constant with respect to $x$ , the derivative of $15 x^{4}$ with respect to $x$ is $15 \frac{d}{d x} [x^{4}]$ .

$f'' (x) = 15 \frac{d}{d x} (x^{4}) + \frac{d}{d x} (- 3 x^{2}) + \frac{d}{d x} (4)$

Step 2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f'' (x) = 15 (4 x^{3}) + \frac{d}{d x} (- 3 x^{2}) + \frac{d}{d x} (4)$

Step 2.2.3

Multiply $4$ by $15$ .

$f'' (x) = 60 x^{3} + \frac{d}{d x} (- 3 x^{2}) + \frac{d}{d x} (4)$

Step 2.3

Evaluate $\frac{d}{d x} [- 3 x^{2}]$ .

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Step 2.3.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x^{2}$ with respect to $x$ is $- 3 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 60 x^{3} - 3 \frac{d}{d x} x^{2} + \frac{d}{d x} (4)$

Step 2.3.2

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

$f'' (x) = 60 x^{3} - 3 (2 x) + \frac{d}{d x} (4)$

Step 2.3.3

Multiply $2$ by $- 3$ .

$f'' (x) = 60 x^{3} - 6 x + \frac{d}{d x} (4)$

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$f'' (x) = 60 x^{3} - 6 x + 0$

Step 2.4.2

Add $60 x^{3} - 6 x$ and $0$ .

$f'' (x) = 60 x^{3} - 6 x$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$15 x^{4} - 3 x^{2} + 4 = 0$

Step 4

Since there is no value of $x$ that makes the first derivative equal to $0$ , there are no local extrema.

No Local Extrema

Step 5

No Local Extrema

Step 6