Calculus Examples

$y = x^{\tan (x)}$ , $x = 1$

Step 1

Find the corresponding $y$ -value to $x = 1$ .

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Step 1.1

Substitute $1$ in for $x$ .

$y = {(1)}^{\tan (1)}$

Step 1.2

Solve for $y$ .

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Step 1.2.1

Remove parentheses.

$y = 1^{\tan (1)}$

Step 1.2.2

Remove parentheses.

$y = {(1)}^{\tan (1)}$

Step 1.2.3

Simplify ${(1)}^{\tan (1)}$ .

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Step 1.2.3.1

Evaluate $\tan (1)$ .

$y = 1^{0.01745506}$

Step 1.2.3.2

One to any power is one.

$y = 1$

Step 2

Find the first derivative and evaluate at $x = 1$ and $y = 1$ to find the slope of the tangent line.

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Step 2.1

Use the properties of logarithms to simplify the differentiation.

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Step 2.1.1

Rewrite $x^{\tan (x)}$ as $e^{\ln (x^{\tan (x)})}$ .

$\frac{d}{d x} [e^{\ln (x^{\tan (x)})}]$

Step 2.1.2

Expand $\ln (x^{\tan (x)})$ by moving $\tan (x)$ outside the logarithm.

$\frac{d}{d x} [e^{\tan (x) \ln (x)}]$

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 $f (x) = e^{x}$ and $g (x) = \tan (x) \ln (x)$ .

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Step 2.2.1

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

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [\tan (x) \ln (x)]$

Step 2.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$e^{u} \frac{d}{d x} [\tan (x) \ln (x)]$

Step 2.2.3

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

$e^{\tan (x) \ln (x)} \frac{d}{d x} [\tan (x) \ln (x)]$

Step 2.3

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = \tan (x)$ and $g (x) = \ln (x)$ .

$e^{\tan (x) \ln (x)} (\tan (x) \frac{d}{d x} [\ln (x)] + \ln (x) \frac{d}{d x} [\tan (x)])$

Step 2.4

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

$e^{\tan (x) \ln (x)} (\tan (x) \frac{1}{x} + \ln (x) \frac{d}{d x} [\tan (x)])$

Step 2.5

Combine $\tan (x)$ and $\frac{1}{x}$ .

$e^{\tan (x) \ln (x)} (\frac{\tan (x)}{x} + \ln (x) \frac{d}{d x} [\tan (x)])$

Step 2.6

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

$e^{\tan (x) \ln (x)} (\frac{\tan (x)}{x} + \ln (x) \sec^{2} (x))$

Step 2.7

Simplify.

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Step 2.7.1

Apply the distributive property.

$e^{\tan (x) \ln (x)} \frac{\tan (x)}{x} + e^{\tan (x) \ln (x)} (\ln (x) \sec^{2} (x))$

Step 2.7.2

Combine $e^{\tan (x) \ln (x)}$ and $\frac{\tan (x)}{x}$ .

$\frac{e^{\tan (x) \ln (x)} \tan (x)}{x} + e^{\tan (x) \ln (x)} \ln (x) \sec^{2} (x)$

Step 2.7.3

Reorder terms.

$e^{\tan (x) \ln (x)} \sec^{2} (x) \ln (x) + \frac{e^{\tan (x) \ln (x)} \tan (x)}{x}$

Step 2.8

Evaluate the derivative at $x = 1$ .

$e^{\tan (1) \ln (1)} \sec^{2} (1) \ln (1) + \frac{e^{\tan (1) \ln (1)} \tan (1)}{1}$

Step 2.9

Simplify.

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Step 2.9.1

Simplify each term.

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Step 2.9.1.1

Evaluate $\tan (1)$ .

$e^{0.01745506 \ln (1)} \sec^{2} (1) \ln (1) + \frac{e^{\tan (1) \ln (1)} \tan (1)}{1}$

Step 2.9.1.2

Simplify $0.01745506 \ln (1)$ by moving $0.01745506$ inside the logarithm.

$e^{\ln (1^{0.01745506})} \sec^{2} (1) \ln (1) + \frac{e^{\tan (1) \ln (1)} \tan (1)}{1}$

Step 2.9.1.3

Exponentiation and log are inverse functions.

$1^{0.01745506} \sec^{2} (1) \ln (1) + \frac{e^{\tan (1) \ln (1)} \tan (1)}{1}$

Step 2.9.1.4

One to any power is one.

$1 \sec^{2} (1) \ln (1) + \frac{e^{\tan (1) \ln (1)} \tan (1)}{1}$

Step 2.9.1.5

Multiply $\sec^{2} (1)$ by $1$ .

$\sec^{2} (1) \ln (1) + \frac{e^{\tan (1) \ln (1)} \tan (1)}{1}$

Step 2.9.1.6

Evaluate $\sec (1)$ .

${1.00015232}^{2} \ln (1) + \frac{e^{\tan (1) \ln (1)} \tan (1)}{1}$

Step 2.9.1.7

Raise $1.00015232$ to the power of $2$ .

$1.00030467 \ln (1) + \frac{e^{\tan (1) \ln (1)} \tan (1)}{1}$

Step 2.9.1.8

The natural logarithm of $1$ is $0$ .

$1.00030467 \cdot 0 + \frac{e^{\tan (1) \ln (1)} \tan (1)}{1}$

Step 2.9.1.9

Multiply $1.00030467$ by $0$ .

$0 + \frac{e^{\tan (1) \ln (1)} \tan (1)}{1}$

Step 2.9.1.10

Divide $e^{\tan (1) \ln (1)} \tan (1)$ by $1$ .

$0 + e^{\tan (1) \ln (1)} \tan (1)$

Step 2.9.1.11

Evaluate $\tan (1)$ .

$0 + e^{0.01745506 \ln (1)} \tan (1)$

Step 2.9.1.12

Simplify $0.01745506 \ln (1)$ by moving $0.01745506$ inside the logarithm.

$0 + e^{\ln (1^{0.01745506})} \tan (1)$

Step 2.9.1.13

Exponentiation and log are inverse functions.

$0 + 1^{0.01745506} \tan (1)$

Step 2.9.1.14

One to any power is one.

$0 + 1 \tan (1)$

Step 2.9.1.15

Multiply $\tan (1)$ by $1$ .

$0 + \tan (1)$

Step 2.9.1.16

Evaluate $\tan (1)$ .

$0 + 0.01745506$

Step 2.9.2

Add $0$ and $0.01745506$ .

$0.01745506$

Step 3

Plug the slope and point values into the point-slope formula and solve for $y$ .

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Step 3.1

Use the slope $0.01745506$ and a given point $(1, 1)$ to substitute for $x_{1}$ and $y_{1}$ in the point-slope form $y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (1) = 0.01745506 \cdot (x - (1))$

Step 3.2

Simplify the equation and keep it in point-slope form.

$y - 1 = 0.01745506 \cdot (x - 1)$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Simplify $0.01745506 \cdot (x - 1)$ .

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Step 3.3.1.1

Rewrite.

$y - 1 = 0 + 0 + 0.01745506 \cdot (x - 1)$

Step 3.3.1.2

Simplify by adding zeros.

$y - 1 = 0.01745506 \cdot (x - 1)$

Step 3.3.1.3

Apply the distributive property.

$y - 1 = 0.01745506 x + 0.01745506 \cdot - 1$

Step 3.3.1.4

Multiply $0.01745506$ by $- 1$ .

$y - 1 = 0.01745506 x - 0.01745506$

Step 3.3.2

Move all terms not containing $y$ to the right side of the equation.

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Step 3.3.2.1

Add $1$ to both sides of the equation.

$y = 0.01745506 x - 0.01745506 + 1$

Step 3.3.2.2

Add $- 0.01745506$ and $1$ .

$y = 0.01745506 x + 0.98254493$

Step 4