Calculus Examples

y = \frac{1}{2} tan (2 x)

$y = \frac{1}{2} \tan (2 x)$

Step 1

Find the first derivative of the function.

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Step 1.1

Since

\frac{1}{2}

$\frac{1}{2}$ is constant with respect to

x

$x$ , the derivative of

\frac{1}{2} \cdot tan (2 x)

$\frac{1}{2} \cdot \tan (2 x)$ with respect to

x

$x$ is

\frac{1}{2} \frac{d}{d x} [tan (2 x)]

$\frac{1}{2} \frac{d}{d x} [\tan (2 x)]$ .

\frac{1}{2} \frac{d}{d x} [tan (2 x)]

$\frac{1}{2} \frac{d}{d x} [\tan (2 x)]$

Step 1.2

Differentiate using the chain rule, which states that

\frac{d}{d x} [f (g (x))]

$\frac{d}{d x} [f (g (x))]$ is

$f' (g (x)) g' (x)$ where

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

$g (x) = 2 x$ .

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Step 1.2.1

To apply the Chain Rule, set

$u$ as

$2 x$ .

$\frac{1}{2} (\frac{d}{d u} [\tan (u)] \frac{d}{d x} [2 x])$

Step 1.2.2

The derivative of

$\tan (u)$ with respect to

$u$ is

$\sec^{2} (u)$ .

$\frac{1}{2} (\sec^{2} (u) \frac{d}{d x} [2 x])$

Step 1.2.3

Replace all occurrences of

$u$ with

$2 x$ .

$\frac{1}{2} (\sec^{2} (2 x) \frac{d}{d x} [2 x])$

Step 1.3

Differentiate.

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Step 1.3.1

Combine

$\sec^{2} (2 x)$ and

$\frac{1}{2}$ .

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

Step 1.3.2

Since

$2$ is constant with respect to

$x$ , the derivative of

$2 x$ with respect to

$x$ is

$2 \frac{d}{d x} [x]$ .

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

Step 1.3.3

Simplify terms.

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Step 1.3.3.1

Combine

$2$ and

$\frac{\sec^{2} (2 x)}{2}$ .

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

Step 1.3.3.2

Cancel the common factor of

$2$ .

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Step 1.3.3.2.1

Cancel the common factor.

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

Step 1.3.3.2.2

Divide

$\sec^{2} (2 x)$ by

$1$ .

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

Step 1.3.4

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

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

Step 1.3.5

Multiply

$\sec^{2} (2 x)$ by

$1$ .

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

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

$g (x) = \sec (2 x)$ .

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Step 2.1.1

To apply the Chain Rule, set

$u_{1}$ as

$\sec (2 x)$ .

$f'' (x) = \frac{d}{d u (1)} ({u_{1}}^{2}) \frac{d}{d x} (\sec (2 x))$

Step 2.1.2

Differentiate using the Power Rule which states that

$\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is

$n {u_{1}}^{n - 1}$ where

$n = 2$ .

$f'' (x) = 2 u_{1} \frac{d}{d x} (\sec (2 x))$

Step 2.1.3

Replace all occurrences of

$u_{1}$ with

$\sec (2 x)$ .

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

Step 2.2

Differentiate using the chain rule, which states that

$\frac{d}{d x} [f (g (x))]$ is

$f' (g (x)) g' (x)$ where

$f (x) = \sec (x)$ and

$g (x) = 2 x$ .

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Step 2.2.1

To apply the Chain Rule, set

$u_{2}$ as

$2 x$ .

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

Step 2.2.2

The derivative of

$\sec (u_{2})$ with respect to

$u_{2}$ is

$\sec (u_{2}) \tan (u_{2})$ .

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

Step 2.2.3

Replace all occurrences of

$u_{2}$ with

$2 x$ .

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

Step 2.3

Raise

$\sec (2 x)$ to the power of

$1$ .

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

Step 2.4

Raise

$\sec (2 x)$ to the power of

$1$ .

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

Step 2.5

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = 2 {\sec (2 x)}^{1 + 1} (\tan (2 x) \frac{d}{d x} (2 x))$

Step 2.6

Add

$1$ and

$1$ .

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

Step 2.7

Since

$2$ is constant with respect to

$x$ , the derivative of

$2 x$ with respect to

$x$ is

$2 \frac{d}{d x} [x]$ .

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

Step 2.8

Multiply

$2$ by

$2$ .

$f'' (x) = 4 \sec^{2} (2 x) \tan (2 x) \frac{d}{d x} (x)$

Step 2.9

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

$f'' (x) = 4 \sec^{2} (2 x) \tan (2 x) \cdot 1$

Step 2.10

Multiply

$4$ by

$1$ .

$f'' (x) = 4 \sec^{2} (2 x) \tan (2 x)$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to

$0$ and solve.

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

Step 4

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

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

Step 5

Simplify

$\pm \sqrt{0}$ .

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Step 5.1

Rewrite

$0$ as

$0^{2}$ .

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

Step 5.2

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

$\sec (2 x) = \pm 0$

Step 5.3

Plus or minus

$0$ is

$0$ .

$\sec (2 x) = 0$

Step 6

The range of secant is

$y \leq - 1$ and

$y \geq 1$ . Since

$0$ does not fall in this range, there is no solution.

No solution

Step 7

Evaluate the second derivative at

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

$4 \sec^{2} (2) \tan (2)$

Step 8

Evaluate the second derivative.

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Step 8.1

Evaluate

$\sec (2)$ .

$4 \cdot {1.00060954}^{2} \tan (2)$

Step 8.2

Raise

$1.00060954$ to the power of

$2$ .

$4 \cdot 1.00121946 \tan (2)$

Step 8.3

Multiply

$4$ by

$1.00121946$ .

$4.00487784 \tan (2)$

Step 8.4

Evaluate

$\tan (2)$ .

$4.00487784 \cdot 0.03492076$

Step 8.5

Multiply

$4.00487784$ by

$0.03492076$ .

$0.13985341$

Step 9

$x =$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x =$ is a local minimum

Step 10

These are the local extrema for

$f (x) = \frac{1}{2} \cdot \tan (2 x)$ .

$(, i s a (l o c a l) (m i n i m u m))$ is a local minima

Step 11