Find the Tangent Line at (0,4) y=4e^xcos(x) , (0,4)

$y = 4 e^{x} \cos (x)$ , $(0, 4)$

Find $\frac{d y}{d x}$ and evaluate at $x = 0$ and $y = 4$ to find the slope of the tangent line at $x = 0$ and $y = 4$ .

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Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} (4 e^{x} \cos (x))$

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Differentiate the right side of the equation.

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Since $4$ is constant with respect to $x$ , the derivative of $4 e^{x} \cos (x)$ with respect to $x$ is $4 \frac{d}{d x} [e^{x} \cos (x)]$ .

$4 \frac{d}{d x} [e^{x} \cos (x)]$

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

$4 (e^{x} \frac{d}{d x} [\cos (x)] + \cos (x) \frac{d}{d x} [e^{x}])$

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

$4 (e^{x} (- \sin (x)) + \cos (x) \frac{d}{d x} [e^{x}])$

Simplify the expression.

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Move $- 1$ to the left of $e^{x}$ .

$4 (- 1 \cdot e^{x} \sin (x) + \cos (x) \frac{d}{d x} [e^{x}])$

Rewrite $- 1 e^{x}$ as $- e^{x}$ .

$4 (- e^{x} \sin (x) + \cos (x) \frac{d}{d x} [e^{x}])$

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

$4 (- e^{x} \sin (x) + \cos (x) e^{x})$

Simplify.

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Apply the distributive property.

$4 (- e^{x} \sin (x)) + 4 (\cos (x) e^{x})$

Multiply $- 1$ by $4$ .

$- 4 e^{x} \sin (x) + 4 \cos (x) e^{x}$

Reorder terms.

$- 4 e^{x} \sin (x) + 4 e^{x} \cos (x)$

Reform the equation by setting the left side equal to the right side.

$y' = - 4 e^{x} \sin (x) + 4 e^{x} \cos (x)$

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = - 4 e^{x} \sin (x) + 4 e^{x} \cos (x)$

Evaluate at $x = 0$ and $y = 4$ .

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Replace the variable $x$ with $0$ in the expression.

$- 4 e^{0} \sin (0) + 4 e^{0} \cos (0)$

Simplify each term.

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Anything raised to $0$ is $1$ .

$- 4 \cdot 1 \sin (0) + 4 e^{0} \cos (0)$

Multiply $- 4$ by $1$ .

$- 4 \sin (0) + 4 e^{0} \cos (0)$

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

$- 4 \cdot 0 + 4 e^{0} \cos (0)$

Multiply $- 4$ by $0$ .

$0 + 4 e^{0} \cos (0)$

Anything raised to $0$ is $1$ .

$0 + 4 \cdot 1 \cos (0)$

Multiply $4$ by $1$ .

$0 + 4 \cos (0)$

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

$0 + 4 \cdot 1$

Multiply $4$ by $1$ .

$0 + 4$

Add $0$ and $4$ .

$4$

Plug in the slope of the tangent line and the $x$ and $y$ values of the point into the point–slope formula $y - y_{1} = m (x - x_{1})$ .

$y - (4) = 4 \cdot (x - (0))$

Simplify.

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The slope-intercept form is $y = m x + b$ , where $m$ is the slope and $b$ is the y-intercept.

$y = m x + b$

Rewrite in slope-intercept form.

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Multiply $- 1$ by $4$ .

$y - 4 = 4 \cdot (x - (0))$

Subtract $0$ from $x$ .

$y - 4 = 4 x$

Add $4$ to both sides of the equation.

$y = 4 x + 4$

Find the Tangent Line at (0,4) y=4e^xcos(x) , (0,4)

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