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Find $h^{\prime}(x)$ using the rules for differentiation:
$h(x)=2 x^{2}+4 x-1$
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The rules for differentiation:
$\frac{d}{d x} a x^{n}=(\text { an }) x^{n-1}$
In words: for each term we multiply the coefficient by the exponent and subtract one from the exponent.
Let's consider the first term of the function and apply the rule:

\begin{aligned} \frac{d}{d x}\left(2 x^{2}\right) &=2(2) x^{(2-1)} \\ &=4 x^{1} \end{aligned}
We repeat this process for each of the remaining terms that make up the given function.
$h^{\prime}(x)=4 x+4$
Important: the derivative of a constant term, such as $-3$, will always be zero.
We can apply the rule to see why this is true:
\begin{aligned} \frac{d}{d x}[-3] &=\frac{d}{d x}\left[-3 \cdot x^{0}\right] \\ &=0 \cdot\left[-3 \cdot x^{0-1}\right] \quad \text { multiply by the } \\ &=0 \cdot\left[\frac{-3}{x}\right] \\ &=0 \end{aligned}
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