Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{1}{2} \cdot \tan (2 x)$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [\tan (2 x)]$ .

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

Step 1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ 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