Calculus Examples

$f (x) = \tan (x) - x$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $\tan (x) - x$ with respect to $x$ is $\frac{d}{d x} [\tan (x)] + \frac{d}{d x} [- x]$ .

$\frac{d}{d x} [\tan (x)] + \frac{d}{d x} [- x]$

Step 1.2

The derivative of $\tan (x)$ with respect to $x$ is $\sec^{2} (x)$ .

$\sec^{2} (x) + \frac{d}{d x} [- x]$

Step 1.3

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

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Step 1.3.1

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

$\sec^{2} (x) - \frac{d}{d x} [x]$

Step 1.3.2

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

$\sec^{2} (x) - 1 \cdot 1$

Step 1.3.3

Multiply $- 1$ by $1$ .

$\sec^{2} (x) - 1$

Step 1.4

Simplify.

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Step 1.4.1

Reorder terms.

$- 1 + \sec^{2} (x)$

Step 1.4.2

Reorder $- 1$ and $\sec^{2} (x)$ .

$\sec^{2} (x) - 1$

Step 1.4.3

Apply pythagorean identity.

$\tan^{2} (x)$

Step 2

Find the second derivative of the function.

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Step 2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \tan (x)$ .

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Step 2.1.1

To apply the Chain Rule, set $u$ as $\tan (x)$ .

$f'' (x) = \frac{d}{d u} (u^{2}) \frac{d}{d x} (\tan (x))$

Step 2.1.2

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

$f'' (x) = 2 u \frac{d}{d x} (\tan (x))$

Step 2.1.3

Replace all occurrences of $u$ with $\tan (x)$ .

$f'' (x) = 2 \tan (x) \frac{d}{d x} (\tan (x))$

Step 2.2

The derivative of $\tan (x)$ with respect to $x$ is $\sec^{2} (x)$ .

$f'' (x) = 2 \tan (x) \sec^{2} (x)$

Step 2.3

Reorder the factors of $2 \tan (x) \sec^{2} (x)$ .

$f'' (x) = 2 \sec^{2} (x) \tan (x)$

Step 3

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

$\tan^{2} (x) = 0$

Step 4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$\tan (x) = \pm \sqrt{0}$

Step 5

Simplify $\pm \sqrt{0}$ .

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Step 5.1

Rewrite $0$ as $0^{2}$ .

$\tan (x) = \pm \sqrt{0^{2}}$

Step 5.2

Pull terms out from under the radical, assuming positive real numbers.

$\tan (x) = \pm 0$

Step 5.3

Plus or minus $0$ is $0$ .

$\tan (x) = 0$

Step 6

Take the inverse tangent of both sides of the equation to extract $x$ from inside the tangent.

$x = \arctan (0)$

Step 7

Simplify the right side.

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Step 7.1

The exact value of $\arctan (0)$ is $0$ .

$x = 0$

Step 8

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$x = π + 0$

Step 9

Add $π$ and $0$ .

$x = π$

Step 10

The solution to the equation $x = 0$ .

$x = 0, π$

Step 11

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$2 \sec^{2} (0) \tan (0)$

Step 12

Evaluate the second derivative.

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Step 12.1

The exact value of $\sec (0)$ is $1$ .

$2 \cdot 1^{2} \tan (0)$

Step 12.2

One to any power is one.

$2 \cdot 1 \tan (0)$

Step 12.3

Multiply $2$ by $1$ .

$2 \tan (0)$

Step 12.4

The exact value of $\tan (0)$ is $0$ .

$2 \cdot 0$

Step 12.5

Multiply $2$ by $0$ .

$0$

Step 13

Since the first derivative test failed, there are no local extrema.

No Local Extrema

Step 14