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Find $D_{x} h(x)$ using the rules for differentiation:
$h(x)=6 x^{4}-6 x^{3}-5 x^{2}$
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The rules for differentiation:
$D_{x} a x^{n}=(a n) x^{n-1}$
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} D_{x}\left(6 x^{4}\right) &=4(6) x^{(4-1)} \\ &=24 x^{3} \end{aligned}
We repeat this process for each of the remaining terms that make up the given function.
$D_{x} h(x)=24 x^{3}-18 x^{2}-10 x$
Important: the derivative of a constant term, such as 4 , will always be zero.
We can apply the rule to see why this is true:
\begin{aligned} D_{x}[4] &=D_{x}\left[4 \cdot x^{0}\right] \\ &=0 \cdot\left[4 \cdot x^{0-1}\right] \quad \longleftrightarrow \quad \begin{array}{l} \text { multiply by the } \\ \text { exponent (zero) } \end{array} \\ &=0 \cdot\left[\frac{4}{x}\right] \\ &=0 \end{aligned}
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