Calculus Examples

y = x^{cos (x)}

$y = x^{\cos (x)}$ ;

x = 1

$x = 1$

Step 1

Find the corresponding

y

$y$ -value to

x = 1

$x = 1$ .

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Step 1.1

Substitute

1

$1$ in for

x

$x$ .

y = {(1)}^{cos (1)}

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

Step 1.2

Solve for

y

$y$ .

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Step 1.2.1

Remove parentheses.

y = 1^{cos (1)}

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

Step 1.2.2

Remove parentheses.

y = {(1)}^{cos (1)}

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

Step 1.2.3

Simplify

{(1)}^{cos (1)}

${(1)}^{\cos (1)}$ .

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Step 1.2.3.1

Evaluate

cos (1)

$\cos (1)$ .

y = 1^{0.99984769}

$y = 1^{0.99984769}$

Step 1.2.3.2

One to any power is one.

y = 1

$y = 1$

y = 1

$y = 1$

y = 1

$y = 1$

y = 1

$y = 1$

Step 2

Find the first derivative and evaluate at

x = 1

$x = 1$ and

y = 1

$y = 1$ 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^{cos (x)}

$x^{\cos (x)}$ as

e^{ln (x^{cos (x)})}

$e^{\ln (x^{\cos (x)})}$ .

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

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

Step 2.1.2

Expand

ln (x^{cos (x)})

$\ln (x^{\cos (x)})$ by moving

cos (x)

$\cos (x)$ outside the logarithm.

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

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

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

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

Step 2.2

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

$f (x) = e^{x}$ and

$g (x) = \cos (x) \ln (x)$ .

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Step 2.2.1

To apply the Chain Rule, set

$u$ as

$\cos (x) \ln (x)$ .

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

Step 2.2.2

Differentiate using the Exponential Rule which states that

$\frac{d}{d u} [a^{u}]$ is

$a^{u} \ln (a)$ where

$a$ =

$e$ .

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

Step 2.2.3

Replace all occurrences of

$u$ with

$\cos (x) \ln (x)$ .

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

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

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

$g (x) = \ln (x)$ .

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

Step 2.4

The derivative of

$\ln (x)$ with respect to

$x$ is

$\frac{1}{x}$ .

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

Step 2.5

Combine

$\cos (x)$ and

$\frac{1}{x}$ .

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

Step 2.6

The derivative of

$\cos (x)$ with respect to

$x$ is

$- \sin (x)$ .

$e^{\cos (x) \ln (x)} (\frac{\cos (x)}{x} + \ln (x) (- \sin (x)))$

Step 2.7

Simplify.

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Step 2.7.1

Apply the distributive property.

$e^{\cos (x) \ln (x)} \frac{\cos (x)}{x} + e^{\cos (x) \ln (x)} (\ln (x) (- \sin (x)))$

Step 2.7.2

Combine

$e^{\cos (x) \ln (x)}$ and

$\frac{\cos (x)}{x}$ .

$\frac{e^{\cos (x) \ln (x)} \cos (x)}{x} + e^{\cos (x) \ln (x)} \ln (x) (- \sin (x))$

Step 2.7.3

Reorder terms.

$- e^{\cos (x) \ln (x)} \sin (x) \ln (x) + \frac{e^{\cos (x) \ln (x)} \cos (x)}{x}$

Step 2.8

Evaluate the derivative at

$x = 1$ .

$- e^{\cos (1) \ln (1)} \sin (1) \ln (1) + \frac{e^{\cos (1) \ln (1)} \cos (1)}{1}$

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

$\cos (1)$ .

$- e^{0.99984769 \ln (1)} \sin (1) \ln (1) + \frac{e^{\cos (1) \ln (1)} \cos (1)}{1}$

Step 2.9.1.2

Simplify

$0.99984769 \ln (1)$ by moving

$0.99984769$ inside the logarithm.

$- e^{\ln (1^{0.99984769})} \sin (1) \ln (1) + \frac{e^{\cos (1) \ln (1)} \cos (1)}{1}$

Step 2.9.1.3

Exponentiation and log are inverse functions.

$- 1^{0.99984769} \sin (1) \ln (1) + \frac{e^{\cos (1) \ln (1)} \cos (1)}{1}$

Step 2.9.1.4

One to any power is one.

$- 1 \cdot 1 \sin (1) \ln (1) + \frac{e^{\cos (1) \ln (1)} \cos (1)}{1}$

Step 2.9.1.5

Multiply

$- 1$ by

$1$ .

$- 1 \sin (1) \ln (1) + \frac{e^{\cos (1) \ln (1)} \cos (1)}{1}$

Step 2.9.1.6

Evaluate

$\sin (1)$ .

$- 1 \cdot 0.0174524 \ln (1) + \frac{e^{\cos (1) \ln (1)} \cos (1)}{1}$

Step 2.9.1.7

Multiply

$- 1$ by

$0.0174524$ .

$- 0.0174524 \ln (1) + \frac{e^{\cos (1) \ln (1)} \cos (1)}{1}$

Step 2.9.1.8

The natural logarithm of

$1$ is

$0$ .

$- 0.0174524 \cdot 0 + \frac{e^{\cos (1) \ln (1)} \cos (1)}{1}$

Step 2.9.1.9

Multiply

$- 0.0174524$ by

$0$ .

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

Step 2.9.1.10

Divide

$e^{\cos (1) \ln (1)} \cos (1)$ by

$1$ .

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

Step 2.9.1.11

Evaluate

$\cos (1)$ .

$0 + e^{0.99984769 \ln (1)} \cos (1)$

Step 2.9.1.12

Simplify

$0.99984769 \ln (1)$ by moving

$0.99984769$ inside the logarithm.

$0 + e^{\ln (1^{0.99984769})} \cos (1)$

Step 2.9.1.13

Exponentiation and log are inverse functions.

$0 + 1^{0.99984769} \cos (1)$

Step 2.9.1.14

One to any power is one.

$0 + 1 \cos (1)$

Step 2.9.1.15

Multiply

$\cos (1)$ by

$1$ .

$0 + \cos (1)$

Step 2.9.1.16

Evaluate

$\cos (1)$ .

$0 + 0.99984769$

Step 2.9.2

Add

$0$ and

$0.99984769$ .

$0.99984769$

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

$0.99984769$ 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

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (1) = 0.99984769 \cdot (x - (1))$

Step 3.2

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

$y - 1 = 0.99984769 \cdot (x - 1)$

Step 3.3

Solve for

$y$ .

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Step 3.3.1

Simplify

$0.99984769 \cdot (x - 1)$ .

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Step 3.3.1.1

Rewrite.

$y - 1 = 0 + 0 + 0.99984769 \cdot (x - 1)$

Step 3.3.1.2

Simplify by adding zeros.

$y - 1 = 0.99984769 \cdot (x - 1)$

Step 3.3.1.3

Apply the distributive property.

$y - 1 = 0.99984769 x + 0.99984769 \cdot - 1$

Step 3.3.1.4

Multiply

$0.99984769$ by

$- 1$ .

$y - 1 = 0.99984769 x - 0.99984769$

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.

$y = 0.99984769 x - 0.99984769 + 1$

Step 3.3.2.2

Add

$- 0.99984769$ and

$1$ .

$y = 0.99984769 x + 0.0001523$

Step 4