# Find the equation of the tangent line to the curve y=x^4+2 e^x at the point (0,2)?

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#### Explanation:

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Feb 10, 2018

$y = 2 x + 2$

#### Explanation:

Now, remember these rules:
$f \left(x\right) = g \left(x\right) + h \left(x\right) \implies f ' \left(x\right) = g ' \left(x\right) + h ' \left(x\right)$

$\frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$

$\frac{d}{\mathrm{dx}} \left(n x\right) = n \cdot \frac{d}{\mathrm{dx}} \left(x\right)$ where $n$ is a constant.

$\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$ where $n$ is a constant.

It is important to keep in mind that $x$ is the variable and $n$ is a constant!

$y = {x}^{4} + 2 {e}^{x}$
=>$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{d}{\mathrm{dx}} \left({x}^{4}\right) + 2 \cdot \frac{d}{\mathrm{dx}} \left({e}^{x}\right)$

=>$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 {x}^{3} + 2 {e}^{x}$

We now plug in $0$ in the place of $x$.

=>$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 \cdot {0}^{3} + 2 {e}^{0}$

=>$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 \cdot {0}^{3} + 2 {e}^{0}$

=>$\frac{\mathrm{dy}}{\mathrm{dx}} = 0 + 2 \cdot 1$

=>$\frac{\mathrm{dy}}{\mathrm{dx}} = 2$

Now, we need to write the equation of the slope.

We use the following information: $y - {y}_{1} = m \left(x - {x}_{1}\right)$

We have: $y - 2 = 2 \left(x - 0\right)$

=>$y = 2 x + 2$

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