Question

Find \(D_{x} h(x)\) using the rules for differentiation:
\[
h(x)=6 x^{4}-6 x^{3}-5 x^{2}
\]

Answer 1

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.
Therefore, the final answer is:
\[
D_{x} h(x)=24 x^{3}-18 x^{2}-10 x
\]
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:
\[
\begin{aligned}
D_{x}[4] &=D_{x}\left[4 \cdot x^{0}\right] \\
&=0 \cdot\left[4 \cdot x^{0-1}\right] \quad \longleftrightarrow \quad \begin{array}{l}
\text { multiply by the } \\
\text { exponent (zero) }
\end{array} \\
&=0 \cdot\left[\frac{4}{x}\right] \\
&=0
\end{aligned}
\]