Calculus Examples

$y = e^{x} \cos (x) + \sin (x)$ , $(0, 1)$

Step 1

Find the first derivative and evaluate at $x = 0$ and $y = 1$ to find the slope of the tangent line.

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Step 1.1

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

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

Step 1.2

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

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Step 1.2.1

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

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

Step 1.2.2

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

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

Step 1.2.3

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

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

Step 1.3

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

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

Step 1.4

Reorder terms.

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

Step 1.5

Evaluate the derivative at $x = 0$ .

$- e^{0} \sin (0) + e^{0} \cos (0) + \cos (0)$

Step 1.6

Simplify.

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Step 1.6.1

Simplify each term.

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Step 1.6.1.1

Anything raised to $0$ is $1$ .

$- 1 \cdot 1 \sin (0) + e^{0} \cos (0) + \cos (0)$

Step 1.6.1.2

Multiply $- 1$ by $1$ .

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

Step 1.6.1.3

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

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

Step 1.6.1.4

Multiply $- 1$ by $0$ .

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

Step 1.6.1.5

Anything raised to $0$ is $1$ .

$0 + 1 \cos (0) + \cos (0)$

Step 1.6.1.6

Multiply $\cos (0)$ by $1$ .

$0 + \cos (0) + \cos (0)$

Step 1.6.1.7

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

$0 + 1 + \cos (0)$

Step 1.6.1.8

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

$0 + 1 + 1$

Step 1.6.2

Simplify by adding numbers.

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Step 1.6.2.1

Add $0$ and $1$ .

$1 + 1$

Step 1.6.2.2

Add $1$ and $1$ .

$2$

Step 2

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

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Step 2.1

Use the slope $2$ and a given point $(0, 1)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (1) = 2 \cdot (x - (0))$

Step 2.2

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

$y - 1 = 2 \cdot (x + 0)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Add $x$ and $0$ .

$y - 1 = 2 x$

Step 2.3.2

Add $1$ to both sides of the equation.

$y = 2 x + 1$

Step 3